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ELEMENTARY MATHEMATICAL ANALYSIS
A SERIES OF MATHEMATICAL TEXTS
EDITED BY
EARLE RAYMOND HEDRICK
THE CALCULUS
By Ellery Williams Davis and William Charles
Brenke.
ANALYTIC GEOMETRY AND ALGEBRA
By Alexander Ziwet and Louis Allen Hopkins.
ELEMENTS OF ANALYTIC GEOMETRY
By Alexander Ziwet and Louis Allen Hopkins.
PLANE AND SPHERICAL TRIGONOMETRY WITH
COMPLETE TABLES
By Alfred Monroe Kenyon and Louis Ingold.
PLANE AND SPHERICAL TRIGONOMETRY WITH
BRIEF TABLES
By Alfred Monroe Kenyon and Louis Ingold.
ELEMENTARY MATHEMATICAL ANALYSIS
By John Wesley Young and Frank Millett Morgan.
COLLEGE ALGEBRA
By Ernest Brown Skinner.
PLANE TRIGONOMETRY FOR SCHOOLS AND COL
LEGES
By Alfred Monroe Kenyon and Louis Ingold.
THE MACMILLAN TABLES
Prepared under the direction of Earle Katmond Hedrick.
PLANE GEOMETRY
By Walter Burton Ford and Charles Ammerman.
PLANE AND SOLID GEOMETRY
By Walter Burton Ford and Charles Ammerman.
SOLID GEOMETRY
By Walter Burton Ford and Charles Ammerman.
CONSTRUCTIVE GEOMETRY
Prepared under the direction of Earle Raymond Hedrick.
JUNIOR HIGH SCHOOL MATHEMATICS
By W. L. VoSBURGH and F. W. Gentleman.
ELEMENTARY MATHEMATICAL
ANALYSIS
BY
JOHN WESLEY YOUNG
PROFESSOR OP MATHEMATICS, DARTMOUTH COLLEGE
AND
FRANK MILLETT MORGAN
ASSISTANT PROFESSOR OF MATHEMATICS, DARTMOUTH COLLEGE
THE MACMILLAN COMPANY
1917
All rights reserved
Copyright, 1917,
By the MACMILLAN COMPANY.
Set up and electrotyped. Published July, 1917.
Reprinted September, October, 191 7.
NoriDoat) }fixtt»
J. 8. Gushing Co. — Berwick & Smith Co.
Norwood, Mass., U.S.A.
PREFACE
This book aims to present a course suitable for students in
the first year of our colleges, universities, and technical schools.
It presupposes on the part of the student only the usual mini
mum entrance requirements in elementary algebra and plane
geometry^
The .book has been written with the hope of contributing
something toward the solution of the problem of increasing
the value and significance of our freshman courses. The
recent widespread discussion of this problem has led to the
general acceptance on the part of many teachers of certain
principles governing the selection and arrangement of mate
rial and the point of view from which it is to be presented.
Among such principles, which have guided us in the prepara
tion of this text, are the following.
1. More emphasis should be placed on insight and under
standing of fundamental conceptions and modes of thought,
less emphasis on algebraic technique and facility of manipula
tion. The development of proficiency in algebraic manipulation
as such we believe has little general educational value. It is
valuable only as a means to an end, not as an end in itself. A
certain amount of skill in algebraic reduction is, of course,
essential to any effective understanding of mathematical pro
cesses, and this minimum of skill the student must secure.
But it seems undesirable in the first year to emphasize the
formal aspects of mathematics beyond what is necessary for
the understanding of mathematical thought. This is espe
cially true for that great majority of students who do not
continue their study of mathematics beyond their freshman
:i79f40ft
vi PREFACE
year and who study mathematics for general cultural and dis
ciplinary purposes. It seems to us altogether probable, how
ever, that even in the case of students who expect to use
mathematics in their later life work (as scientists, engineers,
etc.) greater power will be gained in the same length of time,
if their first year in college is devoted primarily to the gain
ing of insight and appreciation, rather than technical facility.
Experience has shown only too conclusively that in many cases
the emphasis usually placed on formal manipulation effectually
prevents the development of any adequate sort of independent
power.
2. The reference above to the general cultural and disciplin
ary aims of mathematical study at once raises the question as
to the selection of the material that is to form the content of
the course. The cultural motive for the study of mathematics
is found in the fact that mathematics has played and contin
ues to play in increasing measure an important role in human
progress. An educated man or woman should have some con
ception of what mathematics has done and is doing for man
kind and some appreciation of its power and beauty. To this
end our introductory courses should cover as broad a range of
mathematical concepts and processes as possible. In particu
lar, they must not confine themselves to ancient and medieval
mathematics, but must give due consideration to more modern
mathematical disciplines. The fundamental conceptions of
the calculus must be introduced as early as is feasible in view
of the essential role they have played in the progress of
civilization.
If this broad cultural aim is accepted as one of the funda
mental principles in the selection of material, we shall readily
agree that much that is now generally considered necessary
can and should be eliminated from our general courses in
PREFACE vii
mathematics. Almost all of the conventional course in solid
geometry would fall in this category, as well as much of what
is now taught as college algebra, all of the more specialized
portions of analytic geometry, etc. The time thus gained could
then be used for topics that are culturally more significant.
3. The disciplinary motive for the study of mathematics
is the one most often emphasized and ,need not be elaborated
here. In spite of much recent criticism of the doctrine of
formal discipline in education and in spite of the fact that
some of this criticism as applied to mathematics seems to us
justified, we firmly believe that faith in the disciplinary value
of mathematics is fundamentally sound. Teachers of mathe
matics need, however, to formulate with precision their aims
and purposes in this direction and make their practice conform
to this formulation. The disciplinary value of mathematics
is to be sought primarily in the domain of thinking, reasoning,
reflection, analysis ; not in the field of memory, nor of skill
in a highly specialized form of activity. We come back here
to the conflict between insight and technique discussed earlier
in this preface. Suflice it to remark here that the purpose
of technical facility is to economize thought, rather than to
stimulate it. If our primary purpose is to stimulate thought,
we must place the major emphasis on the mathematical formu
lation of a problem and on the interpretation of the final re
sult, rather than on the formal manipulation which forms the
necessary intermediate step ; on the derivation of a formula
rather than merely on its formal application ; on the general
significance of a concept rather than on its specialized function
in a purely mathematical relation.
If we desire to enhance the general disciplinary value of
mathematics, we will seek out and emphasize especially those
conceptions and those modes of thought of our subject which
viii PREFACE
are most general in their application to the problems of our
everyday life. It is fortunate for our purpose — and it is
probably more than a mere coincidence — that the conceptions
and processes of mathematics which most readily suggest
themselves in this connection are the same that are suggested
by the cultural motive discussed earlier. The concept of funo
tionality and the mathematical processes developed for the
study of functions are precisely the things in mathematics
that have most effectively contributed to human progress in
more modern times ; and the thinking stimulated by this
concept and these processes is fundamentally similar to the
thought which we are continually applying to our daily prob
lems. "Functional thinking," to use Klein's famous phrase,
is universal. It comes into play when we make the simplest
purchase, as well as when we attempt to analyze the most
complicated interplay of causes and effects.
In the preparation of this text, we have sought to give an
introduction to the elementary mathematical functions, the
concepts connected therewith, the processes necessary to their
study, and their applications. By making the concept of a
function fundamental throughout we believe we have gained
a measure of unity impossible when the year is split up among
several different subjects. The arrangement of this material
is exhibited in the table of contents and the text proper, and
need not be discussed here. We would merely call attention
briefly to some features which seem to require emphasis or
explanation.
The change in the value of a function due to a change in
the value of a variable is emphasized from the very beginning.
The change ratio A y/A x is introduced in Chapter III for the
linear function, and the derivative is introduced as the slope
of the graph of a quadratic function in Chapter IV, although
PREFACE ix
the word "derivative" is not introduced until Chapter XIX.
Derivatives are used in Chapters IV, V, X, XII, XIII, and XIX.
We have discussed rather more fully than is customary
those topics which involve new and important concepts, and
have been correspondingly brief where we felt the student
ought to be able to supply the argument himself. We have
tried throughout to place the emphasis on an understanding
of the general bearing of the principles, and have consistently
tried to minimize difficulties of mere algebraic technique. It
seems quite likely that customary classroom procedure will,
therefore, need to be modified in the direction of lessening the
time given to formal recitations and increasing the opportuni
ties for informal discussion. A number of questions have
been inserted among the exercises which it is hoped will
stimulate such discussion ; this is the purpose also of a num
ber of the " Why's " scattered throughout the text.
The lists of "Miscellaneous Exercises" found at the end
of chapters beginning with Chapter XI contain some exercises
too difficult for assignment in an average class. These may
be used to advantage, we hope, in socalled " honor sections "
consisting of men who have shown exceptional ability in
mathematics.
A word regarding our conception as to how the text may
be applied to meet the varying mathematical preparation of
students will not be out of place. At Dartmouth College we
propose to distinguish in this connection only two kinds of
freshmen : those who enter without trigonometry, and those
who have passed a course in trigonometry in their secondary
school. The former will cover the first fifteen chapters of
this text in a course meeting three hours per week throughout
the year (about ninety assignments). These men will have
all the necessary preliminarj^ training for the usual courses
X PREFACE
in the calculus. Those students who enter with trigonometry
will cover the first nineteen chapters in a course meeting three
hours per week throughout the year, covering the material of
Chapters VI, VII, VIII, and IX (which for them is largely
review) in about three weeks.
In a course meeting five times per week throughout the
year, there should be ample time also for a thorough study
of the important topics of Chapter XX (Determinants) and
Chapters XXIXXII (Functions of two independent variables ;
analytic geometry of space).
So much has been said in recent years in favor of a unified
course in mathematics for freshmen that it seems desirable
actually to try it out in practice. For this purpose a text
book is necessary. We do not believe that this text will
solve the problem ; the most we can hope for is that we have
secured a first approximation. It is for this reason that we
urgently request users of this text to communicate to us any
criticisms or suggestions that occur to them looking to the
improvement of possible later editions. In particular, we
should like advice and counsel as to the possible desirability
of increasing the amount of calculus included in the first year.
This could be done by devoting less space to the purely geo
metric aspects of analytic geometry. On theoretical grounds
we believe this to be desirable. We felt, however, that we
ought to be conservative in case of an innovation of this sort,
with a view of seeing how the introduction to this limited
extent of the notion of the derivative in the first year fares.
If the results are satisfactory, we could then take the next step
with confidence.
J. W. YOUNG.
F. M. MORGAN.
Hanover, N. H.,
AprU, 1917.
CONTENTS
PART I. INTRODUCTORY CONCEPTIONS
Chapter I. Functions and Their Representation
Chapter II. Algebraic Principles and Their Connection
with Geometry
132
Chapter
Chapter
I.
IT.
III.
Chapter
Chapter
Chapter
Chapter
PART II. ELEMENTARY FUNCTIONS
III. The Linear Function. The Straight Line
IV. The Quadratic Function .
Graphs of Quadratic Functions .
Applications of Quadratic Functions '.
Quadratic Equations ....
V. The Cubic Function. The Function jr"
VI. The Trigonometric Functions
VII. Trigonometric Relations
VIII. The Logarithmic and Exponential
tions
Chapter IX. Numerical Computation .
I. Errors in Computation ....
II. Logarithmic Solution of Triangles
III. The Logarithmic Scale. The Slide Rule
IV. Logarithmic Paper ....
Chapter X. The Implicit Quadratic Functions .
Func
6497
98128
98114
115119
120128
129142
143187
188211
212235
236264
236241
242251
252259
260264
PART III. APPLICATIONS TO GEOMETRY
Chapter XL The Straight Line 293319
Chapter XII. The Circle
xi
xii CONTENTS
Chapter XIII. The Conic Sections . 887876
Chapter XIV. Polar Coordinates . . 877891
Chapter XV. Parametric Equations 892401
PART IV. GENERAL ALGEBRAIC METHODS —
THE GENERAL POLYNOMIAL FUNCTION
Chapter XVI. Miscellaneous Algebraic Methods . . 402419
Chapter XVII. Permutations, Combinations, and Proba
bility. The Binomial Theorem 420431
Chapter XVIII. Complex Numbers 482448
Chapter XIX. The General Polynomial Function. The
Theory of Equations .... 449474
Chapter XX. Determinants 476498
PART V. FUNCTIONS OF TWO VARIABLES
— SOLID ANALYTIC GEOMETRY
Chapter XXI. Linear Functions. The Plane and Straight
Line 494618
Chapter XXII. Quadratic Functions. Quadric Surfaces . 614681
Tables . . . • 684642
Powers and Roots 534
Important Constants 535
FourPlace Logarithms 536537
FourPlace Trigonometric Functions .... 538542
Index 648648
ELEMENTARY MATHEMATICAL ANALYSIS
ELEMENTARY
MATHEMATICAL ANALYSIS
PART I. INTRODUCTORY CONCEPTIONS
CHAPTER I
FUNCTIONS AND THEIR REPRESENTATION
1. The General Idea of a Function. Our daily activities
continually furnish us with examples of things that are related
to one anotherj of quantities which depend on certain other
quantities, which change when certain other quantities change.
Thus, a man's health is related to the food he eats, the exer
cise he takes, and to many other things. The price of any
manufactured article depends on the cost of production, while
the latter cost in turn depends on the cost of the raw ma
terial, the cost of labor, etc. The weather depends on a
variety of conditions. These are complicated examples of
dependence. There are very simple examples. Thus the
price paid for a certain quantity of sugar depends on the num
ber of pounds bought and the price per pound ; the area of a
square depends on the length of one of its sides ; and so forth.
In all such cases, where some quantity depends on some
other quantity or quantities, we say that the former is a func
tion of the latter. Thus the price of an article is a function of
the cost of production, the area of a square is a function of the
length of one of its sides, etc.
B 1
2 MATHKMATICAL ANALYSIS [I, § 2
2. General Laws. Many problems of science consist in
expressing as accurately as possible one quantity in terms of
another quantity on which the first depends. The statements,
" The area of a square is equal to the square of the length of
one side," and " The speed of a body falling from rest is pro
portional to the time it has fallen " are simple examples.
At the basis of this idea of dependence or functionality is
the notion of a general law which the quantities in question
obey. Most of the problems of civilized life are concerned,
directly or indirectly, with the investigation of such laws.
Thus medical science seeks to discover the laws governing
health, economies' investigates the laws governing the produc
tion and distribution of wealth, the business man studies the
conditions which influence his business and his profits. In
every case the investigation of the law in question involves
finding out how something is related to, depends on, changes
with, something else ; i.e. the study of a function of some kind.
The ability to think clearly about such relationships is of
the highest importance to every one. This course in mathe
matics is primarily concerned with the study of certain of the
simpler kinds of functions and their applications.
3. Numbers and Quantities. We shall confine ourselves
in general to the study of relations between things which can
be ^measured. We can then always speak of the amount of one
of them. Such an amount is expressed, in terms of a suitable
unit of measure^ by means of a number. Anything that can be
represented by means of a number we shall call a quantity.
A function expressing the relation of one such quantity to
another gives rise to a relation between numbers. A very power
ful aid in studying functions is their geometric representation,
which we shall discuss presently. We must consider first,
however, the geometric representation of a single quantity.
I, § 5] REPRESENTATION OF FUNCTIONS 3
4. The Arithmetic Scale. The distinction between two of
the simplest kinds of quantities can be illustrated by reference
to their geometric or graphic representation. Every one is
familiar with the socalled arithmetic scale (Fig. 1), of which the
I i j I \ 1 1 ^ 1 . , . ^ r
J 3 3 4 5 6
Fig. 1
yard stick and tape measure are examples. The divisions of
the scale in these cases represent lengths. Another example
is the beam on a certain kind of balance ; here the divisions of
the scale represent weights.
A characteristic feature of an arithmetic scale is that it
begins at some point and extends from in one direction.
The quantities represented by such a scale are expressed by
means of the numbers of arithmetic. These in turn represent
simply the magnitude, or the size, or the amount, of something
(as 12 yd. of cloth, 96 lb. of sugar, etc.).
5. The Algebraic Scale. Hardly less familiar nowadays is
the socalled algebraic scale (Fig. 2). The most familiar ex
n — ' — I — I — I — ' — r^ — I J , I — ' — r
3 2 •^l/ *1 +2i +3
Fia. 2
ample is probably the scale on an ordinary thermometer.
Every one knows the meaning of f 10° or — 5°.
Such an algebraic scale extends in two opposite directions
from some arbitrary point (marked 0) of the scale. The quan
tities represented by the points of such a scale are expressed
by means of the socalled real numbers of algebra, such as :
...,  4,  Vi2,  3,  i 0, + 1, + ii, .
Such a number represents not merely a magnitude, but
rather a magnitude and one of two opposite directions or
4 MATHEMATICAL ANALYSIS [I, § 5
senses. These two opposite " senses " are of various kinds
according to tlie quantities considered. They are often ex
pressed by such phrases as : " to the right of " and " to the
left of," " above " and " below," " greater than " and " less
than," " before " and " after," etc. Thus + 10° of temperature
means a temperature 10° greater than the arbitrary temperature
which we have agreed to indicate by 0° ; whereas — 5° means
a temperature 5° less than the temperature indicated l)y 0°.
It should be noted that 0° of temperature does not mean the
absence of temperature.
6. Magnitudes and Directed Quantities. We have seen
in the last two sections that a number may represent simply a
magnitude; or, that a number may represent a magnitude and
one of two opposite directions. The numbers of arithmetic serve
the former purpose, the positive and negative numbers of
algebra serve the latter. Thus the number 5 represents
simply a magnitude, such as a distance of five miles between
two stations or a period of time of five hours. The numbers
h 5 and — 5 also represent magnitudes of five units ; but
they represent more than this. They may tell us, for example,
that a station is five miles east of a certain place denoted by
and that another station is five miles west of the place denoted
by 0, respectively ; or that an event took place five hours after
or five hours before a certain event.
We may then distinguish two kinds of quantities : (1) mag
nitudes, and (2) socalled directed quaiitities. Examples of the
former are : the length of a board, the weight of a barrel of
flour, the duration of a period of time, the price of a loaf
of bread, etc. Examples of the latter are : the temperature (a
certain number of degrees above or below zero), the distance
and direction of some point yl on a line from some other
point B on the line^ the time at which a certain event
, , ,5 ,
.  """^ , ■ 
< . ."^.
I, § 7] REPRESENTATION OF FUNCTIONS 5
occurred (a certain number of hours before or after a given
instant) ; etc.*
Geometrically, the distinction between directed quantities
and mere magnitudes corresponds to the fact that, on the one
hand, we may think of the line segment AB as drawn from A to
B or from B to A, and, on the other hand, we
may choose to consider only the length of
such a segment, irrespective of its direction.
Figure 3 exhibits the geometric representation
of 5, 4 5, and — 5. A segment whose direc
tion is definitely taken account of is called a directed segment.
The magnitude of a directed quantity is called its absolute
value. Thus the absolute value of — 5 (and also of h 5) is 5.
7. Further Remarks concerning Scales. Scales, both arith
metic and algebraic, occur in practice in a variety of forms. We have
hitherto considered only the simplest form, in M^hich the scale is con
structed on a straight line and in which the subdivisions corresponding
to the numbers 1, 2, 3, ... (and in case of the algebraic scale also those
corresponding to the numbers — 1, —2, —3, •••) are at equal intervals.
Neither of these two conditions is essential. A scale may be constructed
on a curved line (a circle, for example, in which case it is sometimes
called a dial). Scales are also used in which the intervals between the
points representing the whole numbers are not equal. Such a scale is
called a nonuniform scale. The scales on some forms of thermometers,
on a slide rule (see p. 252), on certain types of ammeters and pressure
gauges, etc., may serve as examples of nonuniform scales. The scales
discussed in §§ 4, 5 are then to be described more fully as rectilinear and
uniform. In the future, unless specifically stated otherwise, a scale will
always mean a uniform scale.
* We are here considering only magnitudes in one of two opposite directions
It is also possible to consider as quantities magnitudes taken in any direction
in a plane or in space. Thus a force has a certain magnitude
and is exerted in a certain direction ; it could be completely
represented by a line segment whose length represents the
magnitude of the force and whose direction (shown by arrow
head) represents the direction in which it acts. Such quantities are called
vectors. We shall have occasion to refer to them again (Chap. XVIII) .
6
MATHEMATICAL ANALYSIS
[I, §8
8. Use of Line Segments to Represent Quantities. Statis
tical Data. A common use of line segments to represent
quantities is in connection with the graphic representation of
statistical data. The table below, for example, gives the areas
of the New England States ; the adjacent figure represents
these areas by means of line segments.
Area of New England States
States
Maine
Vermont
New
Hampshire
Massa
chusetts
Connec
ticut
Rhode
Island
Square Miles
33,040
9,565
9,305
8,315
4,990
1,250
Maine
Vermont
New Hamp,
Massachusetts
Connecticut
Rhode Island
JO
15
Fig. 4
?o
25
30 Thous.sq.miles
The method of constructing such a graphic representation
should be clear without further comment.
The above areas could also be represented by areas, as in the following
figure.
Vt.
N.H.
Mass.
E3
Fig. 5
In general, this method of representation is not so serviceable. Why ?
I, § 8] REPRESENTATION OF FUNCTIONS 7
EXERCISES
1. From the following table represent graphically by means of line
segments the enrolment in Dartmouth College during the years 19011916 :
'01'02 '02'03 '03'04 '04'05 '05'06 '06'07 "07'08 '08'09
686 709 802 857 927 1058 1131 1136
'09'10 'lO'll '11'12 '12'13 13' 14 '14' 15 '15'16
1197 1165 1242 1246 1284 1336 1422
Use a convenient unit to represent 100 students (say \ in.). Can you
then represent the data with complete accuracy ? Why ?
2. Represent graphically the size of the libraries of the following
institutions :
No. of Volumes No. of Volumes
Harvard 1,180,000 Williams 80,000
Yale 1,000,000 Amherst 110,000
Dartmouth 130,000 Wesleyan 100,000
Brown 115,000 Univ. of Vermont . . 91,000
3. Take the edge of a sheet of paper and mark on it a point A. Place
this edge along the segment representing the area of Vt. in Fig. 4, the
point A coinciding with the lefthand extremity of the segment. Mark
the righthand extremity by a point B on the paper. Do the same with
the segment representing N. H., placing the point B at the lefthand
extremity, however, and obtaining a new point C, corresponding to the
righthand extremity. Continue this process for the states Mass., Conn.,
and R. I. The total segment represents the sum of the areas. Show that
Me. has an area almost as great as that of the other N. E. states com
bined. The process just described in the above exercise is known as
graphic addition.
4. Describe a similar process for graphic subtraction.
5. Show that the distance between two points of an arithmetic scale
can always be found by subtraction. Is the same true for the points of
an algebraic scale ? What is the meaning of the sign of the difference ?
6. Two algebraic scales intersect at right angles, the point of intersec
tion being the point of each scale, and the units on both scales being
the same. Show how to find the distance from any point on one scale to
any point on the other. Would your method still be applicable, if the
units on the two scales were different ? Explain your answer.
7. In constructing Fig. 5 what theorem of plane geometry regarding
the areas of similar figures is used ? Could the result of Ex. 3 have been
readily obtained from the representation in Fig. 5 ?
8
MATHEMATICAL ANALYSIS
[I, §9
9. The Investigation of Functions. We are now ready
to consider in some detail a few special examples of func
tions, in order to familiarize ourselves with certain gen
eral characteristics a function may possess, with certain
methods for the representation and study of functions, and
with the terminology. This is desirable before taking up the
more systematic study of general types of functions.
10. Example i. Tlie temperature as a function of the time.
The temperature at a given place is a function of the time of
day. At any given time we can determine the temperature by
/ 7 'Wednesday I S TJmrsday J 9 Fn 1<j;/
Fig. 6
simply reading an ordinary thermometer. For the meteorolo
gist, however, the actual temperature at any instant is of less
importance than the changes in the temperature that take place
during a period of time (such as a day, a month, etc.). To
trace these changes he must know the temperature at every
I, § 10] REPRESENTATION OF FUNCTIONS
9
instant. For this purpose he makes use of a selfrecording
thermometer. A portion of a record of such a thermometer is
given in Fig. 6.
The way in which such an instrument works is briefly as follows.
The pivoted lever shown in the figure (Fig. 7) carries a pencil point. The
mechanism of the instrument causes the pencil end of the lever to rise or
fall as the temperature rises or falls, so that if a vertical thermometer
scale* were adjusted behind the pencil point we could read off the
Fig. 7
temperature on this scale. The pencil point rests against a strip of paper,
ruled as in Fig. 6, which is mounted on a drum. Clockwork causes this
drum to rotate uniformly at the proper speed. The rulings on the strip
of paper now explain themselves. The distance between two successive
horizontal lines corresponds to 2° of temperature. The distance between
two successive vertical arcs corresponds to two hours. The temperature
at any instant can then be read from the record on the strip of paper.
The way in which such a record may be used is illustrated
by the following questions, which refer to the record of Fig. 6.
* Since the pencil moves on an arc of a circle, this vertical scale is con
veniently constructed on such an arc, rather than on a straight line.
10 MATHEMATICAL ANALYSIS [I, § 10
1. What was the temperature at noon on each of the three days given ?
2. What was the temperature at midnight between Wednesday and
Thursday ? At 6 p.m. on Friday ?
3. What was the maximum and the minimum temperature on each of
the three days, and at what times did it occur ?
4. When was the temperature 50° ? During what periods was it above
50°?
5. How would a stationary temperature be recorded ? A rapidly
rising temperature ? A rapidly falling temperature ?
6. By how many degrees did the temperature change on Wednesday
from noon to 2 p.m. ? Was this change a rise or a fall ?
7. During what two hours on these three days did the greatest rise in
temperature occur ?
8. When did the most rapid rise in temperature occur ? When the
most rapid fall ?
9. What was the average rate of increase (in degrees per hour) in the
temperature from the minimum on Thursday to the maximum on Thurs
day ? The average rate of decrease from the maximum on Wednesday to
the following minimum ?
11. Graphic Representation. In the preceding example we
exhibited the temperature as a function of the time by means
of a curve drawn with reference to a time scale and a tempera
ture scale. Such a curve is called a graph of the function in
question. Such a graphic representation gives a vivid picture
of the function ; but it is limited in accuracy. Why ? Can a
change in temperature of 0.1° be distinguished on this graph ?
12. Example 2. Speed in terms of the time. Keadings of
the speedometer of an automobile taken every five seconds
from a standing start are given in the following table :
Number of seconds after start 5 10. 15 20 25 30 35
Speed in miles per hour 2 6 7 16 21 28 36
We proceed to construct a graph of the function thus ob
tained, as follows. We take a piece of squareruled paper and
on one of the horizontal lines (which for convenience we draw
more heavily) construct a uniform scale to represent the time
I, § 12] REPRESENTATION OF FUNCTIONS
11
(Fig. 8). On the vertical lines through the points representing
5, 10, 15, 20, . . . seconds we lay off segments to represent the
speeds at the respective instants. This is most conveniently
done by constructing on the vertical line through a scale
representing speed in miles per hour. Thus, by reference to
the scale indicated in the figure, the point A represents the
«y
■ ~

r
~
~
■"
~
~
~
—
"

~
"
"
"
■
^
/
f
^
1/
/
/
/
</n
4
• W
,
/
/
f
f
J
t
1
i
*
.
•X
'
^
'
::
J
_
10 no i
Seconds after start
Speed of an Automobile
Fig. 8
corresponding values : 15 seconds and 7 miles per hour. The
other points indicated in the figure are now readily located, or
" plotted," in similar fashion. The final step in constructing the
figure consists in drawing a " smooth curve " through the points.
The curve thus obtained may be used as was the tempera
ture curve discussed in the previous example. We might, for
example, conclude from this figure that the speed of the car at
the end of 23 seconds was probably about 18 ^ miles per hour.
12 MATHEMATICAL ANALYSIS [I, § 12
The necessity of saying " probably ", however, exhibits an
essential difference between this example and the former one.
In case of the temperature record the temperature at every
instant was automatically recorded ; any point of the curve in
that example was as significant as any other point. In the
present example the only speeds actually measured are those
specifically listed in the above table. And yet the conclusion
stated above regarding the speed of the car at the end of 23
seconds is justified. Why ?
1. What was the probable speed of the car at the end of 27 seconds?
2. How long did it take the car to pick up from to 30 miles per hour ?
3. The driver probably shifted gears between the 10th and 15th seconds.
What can be said of the reliability of the curve during this interval ?
4. How is the steepness of the curve related to the rate at which the
speed is increasing ?
6. Is it possible to calculate, by the use of this figure, approximately
how far the car traveled during the first 35 seconds ?
13. Variables. It is desirable to introduce at this point a
certain terminology. In the preceding examples we have
considered temperature and speed as functions of (i.e. de
pendent on) the time. We have considered several different
instants of time and the corresponding values of the tem
perature and the speed. Whenever, in a given discussion,
we consider a number of different values of a quantity,
such as time, or temperature, or distance, or weight, etc.,
we call such a quantity a variable. In the above examples,
the time and the temperature and the speed are all varia
bles ; and, since in the first example we have thought of
the temperature as depending on the time, we may speak
of the temperature as the dependent variable, of the time
as the independe7it variable. It is often more convenient,
however, to call the dependent variable simply the function
I, § 15] REPRESENTATION OF FUNCTIONS
13
and the independent variable the variable. Thus, in the
second example, the speed was the function and the time
was the variable.
14. Tabular Representation. Interpolation. In the second
example we secured data concerning a function by measurement
and exhibited the corresponding values of variable and function
by means of a table of values. Such a table is called a tabular
representation of the function. The accuracy of such a repre
sentation is limited only by the precision of measurement.
Such a table, however, gives an incomplete description of the
function. AVhy? The process of obtaining values of the
function for values of the variable that lie between the re
corded values stated in the table is called interpolation.
When the interpolated values are read from a graph of the
function, the process is known as graphic interpolation. The
answers to the first two questions at the end of § 12 were
obtained by graphic interpolation.
15. Example 3. Volume of ivater as a function of the tem
perature. When 1000 cc. of water at 0° centigrade is heated,
it is found that the volume of the water changes according to
the following table.
Degrees Centigrade
Cubic Centimeters
1000.00
2
999.90
4
999.87
6
999.90
8
999.98
Degiees Centigrade
Cubic Centimeters
10
1000.12
12
1000.32
14
1000.57
16
1000.86
20
1001.61
It requires a rather careful examination of this table to learn
that as the temperature (the variable) is increased 'from 0°
the volume of the water (the function) decreases and then in
creases. A graphic representation of this function, analogous
to the examples already considered in §§ 10, 12, would have
yielded this result at a glance. It is our next concern to
14
MATHEMATICAL ANALYSIS
[I, § 15
see how such a representation can be constructed, in this
case.
To this end we a take a piece of squareruled paper and on
one of the horizontal lines construct a uniform scale to repre
sent temperatures. At the points representing 0°, 2°, 4°, 6°,
•••, we would then lay off on the vertical lines distances that are
1 1
^


'

r
r
r
■^
f
y
/
1
JJ
/
/
1001.0
f
/
>
'
/
/
/
/
/
/
/
1000.0
r
^
/'
s
«v,
^
f
9
^m
^
s
1
999.0
^ ""* FlO 9 ^ Degrees Centigrade,
to represent the volumes in which we are interested.
However, at this point a difficulty presents itself. The
numbers representing the volumes in question are so large, and
the differences between the volumes for the various tempera
tures so small, that, if we choose the unit on the vertical
scale small enough to represent these volumes on a sheet of
paper of convenient size, it would be a practical impossibility
I, § 15] REPRESENTATION OF FUNCTIONS 15
to represent the volumes with sufficient accuracy to make the
diiferences in the volumes distinguishable. It is precisely
these variations in volume, however, in which we are primarily
interested.
To overcome this difficulty, we adopt the expedient of ex
hibiting merely that portion of the graphic representation in
which we are primarily interested, and are then able to use a
largely magnified scale. That is, we observe that all the
volumes in which we are interested lie between 999.00 cc. and
1001.00 cc. We may then assume that the points on the line
on which we constructed the temperature scale are at a height
representing 999.00 cc. (Fig. 9). In other words we suppose
the zero point' of the vertical volume scale to be a great dis
tance below the point at which we are working. We construct
a portion of the volume scale on the vertical line through 0,
marking the latter point 999.0 and choosing the unit on this
scale sufficiently large to meet our requirements. In the
figure, as drawn, each of the vertical divisions represents
0.1 cc. The construction of the points P, Q, R, ••• is then
readily made. A smooth curve drawn through the points thus
plotted then gives the graph *of the function.
Here, again, the points in the curve between the points
given by the table are uncertain ; but the regularity with
which the given points are arranged together with the nature
of the phenomenon we are considering leaves little room for
doubt that, if the volumes for 1°, 3°, 5°, — should be measured
and the resulting volumes plotted, the resulting points would
be located upon (or at least very near to) the curve drawn.
1; What is the volume of water at 1° ? at 19° ?
2. What is the minimum volume, and at what temperature does it
occur ?
3. At what temperature besides 0° is the volume 1000.00 cc?
16
MATHEMATICAL ANALYSIS
[I, § 15
EXERCISES
1. The following temperatures were observed at Hanover, N.H., on a
certain day in February, 1914 :
Midnight
12°F.
1 A.M.
13°
9 a.m.
 12° F.
6 P.M.
+ 18° F.
2 A.M.
14°
10 A.M.
 2°
6 p.m.
+ 11°
3 a.m.
15°
11 A.M.
+ 4°
7 P.M.
+ 6°
4 A.M.
 17°
Noon
+ 10°
8 p.m.
+ 2°
5 A.M.
20°
1 P.M.
+ 12°
9 P.M.
+ 1°
6 A.M.
21°
2 P.M.
+ 14°
10 p.m.
0°
7 A.M.
22°
3 p.m.
+ 19°
11 p.m.
 2°
8 a.m.
19°
4 p.m.
+ 22°
Midnight
' 4°
Plot the corresponding points on squareruled paper, and draw an
approximate graph of the function. Assuming this graph to be correct,
what was the temperature at 6.30 a.m.? At 6.30 p.m.? What was the
total range (the difference between the maximum and the minimum)
of temperature during the day ? How long did it take the temperature
to rise from its minimum to its maximum ? At what average rate in
degrees per hour did the temperature rise during this period ?
2. A stiff wire spring under tension is found experimentally to stretch an
amount d under a tension T as follows :
r in lb
10
15
20
25
30
d in thousandths of in. .
8 .
12
16.3 '^
20
23.5
Plot the above data. What would the stretch be when the tension is
12 1b.? 271b.? 23 1b.?
3. The intercollegiate track records are as follows, where d is the dis
tance run and t is the time :
d
100 yd.
220 yd.
440 yd.
880 yd.
1 mile
2 miles
t'
9^ sec.
2H sec.
48 sec.
1 m. b^ sec.
4 m. 14f sec.
9 m. 23 sec.
I, § 15] REPRESENTATION OF FUNCTIONS
17
Plot these records by points in a plane, and draw a smooth curve through
them. Are the points of this curve significant ? Why ? What would
you expect the record for 600 yd. to be ? For 1600 yd.? For 1000 yd.?
Compare the results of these interpolations with the actual records for
these distances.
4. The following table shows the distance at which objects at sea
level are visible from certain elevations :
Elevation
Distance
Elevation
Distance
Elevation
Distance
Feet
Miles
Feet
Miles
Feet
Miles
1
1.3
40
8.4
200
18.7
5
3.0
50
9.3
300
22.9
10
4.2
100
13.2
500
29.6
20
5.9
150
16.2
1000
33.4
30
7.2
Plot the graph of this function. Use a different scale for elevation for
values from 100 to 1000 ft. from that used from 1 to 50 ft. Why ?
5. The following is an extract of the mortality table prescribed by
statute in most states as the basis on which the reserves of life insurance
companies shall be computed :
Age
Number
Living
Age
Number
Living
Age
Number
Living
10
15
20
26
* 30
35
100,000
96,285
92,637
89,032
85,441
81,822
40
45
50
55
60
65
78,106
74,173
69,804
64,563
57,917
49,341
70
75
80
85
90
95
38,569
26,237
14,474
5,485
847
3
Draw the mortality curve. Of 100,000 living at the age of 10 years
approximately how many would be alive at 32 years ? At 57 years ?
How would you represent on the graph the number dying during any
given period of five years ?
18
MATHEMATICAL ANALYSIS
[I, § 16
16. Empirical Functions and Arbitrary Functions. The
examples of functions we have hitherto considered have been
taken from observed measurements of relations existing in
nature and life about us. Such functions are called empirical.
Another type of functions may now engage our attention.
They may be called arbitrary or artfjicial The following will
serve as an example.
17. Example 4. Letter postage. According to the postal
regulations the postage on letters is fixed at two cents per
u
u
10
s
e
:
2 3 4
Letter Postage
Fig. 10
G Ounces
ounce or fraction thereof. The graph showing the relation
between the amount of postage and the weight of the letter
is then given by Figure 10.
18. Constant Functions. Continuous and Discontinuous
Functions. The graph just referred to exhibits two peculiar
ities that we have not yet had occasion to observe in connection
with a function.
(1) The value of the function may make a sudden jump as
the variable passes through certain values (in this case when
the weight passes through the values 1 oz., 2 oz., etc.) without
taking on the intermediate values. In the present case, as the
weight is increased from exactly 1 oz. to the slightest amount
I, § 19] REPRESENTATION OF FUNCTIONS 19
above 1 oz. the postage jumps from 2 cents to 4 cents. A
function with such breaks, or changes of a definite amount
for no matter how slight a change in the variable, is said to be
discontinuous for those values of the variable at which the
break or jump occurs.
A function, on the other hand, whose graph is a continuous
line or curve without such sudden breaks or changes is said to
be a continuous function.*
(2) Portions of this graph are horizontal straight lines, which
means that certain changes in the variable produce no corre
sponding change in the value of the function. Thus, the
postage does not change as the weight of the letter is in
creased from slightly more than 1 oz. to 2 oz. In such a case
we say that the function is constant (or stationary) for the
interval of the variable in question.
We should observe, further, that the graph of the function
as drawn does not furnish a unique value for the function at
the points of discontinuity, i.e. when the weight is 1, 2, 3, •••
oz., since there is nothing to indicate whether we should take
the lower or the higher value. As a matter of fact the arbi
trary definition of the function specifies that the lower value is
to be taken.
19. More about Arbitrary Functions. We must not assume,
of course, from the preceding example that every arbitrary
function is discontinuous.
In fact, we should note that if we take any squareruled
paper, construct on it a horizontal scale, any number of which
we will designate by x, and a vertical scale, any number of
* The word continuous is used in mathematics in a highly technical sense,
the full discussion of which is beyond the scope of an elementary course.
The definition of the term given above is sufficiently precise for our present
purposes. Later we shall have more to say of it.
20
MATHEMATICAL ANALYSIS
[I, § 19
which we will call y, and then draw an arbitrary curve across
the paper, as in Fig. 11, we thereby define a relation between
the numbers x of the horizontal scale and the numbers y of
the vertical scale, such that to every value of x corresponds a
certain value (or possibly a set of values) of y ; i.e. we define
y as a function of x.* The .reason for the phrase in paren
5eii

...
 ■ w

" " ' t
''

::
'^
:
ft
:
±:i
L
r
. ! ■,
iij
• 1
is , .
" "'4'"
. .
. . . . •
.
:::
i::
^\'.:::i
:
5
T
i^^ij
>4 ■'
:
ir
T^l
ii
M

±i
;
Fig. 11
Fig. 12
theses in the last sentence is as follows. If the curve we draw
is such that for any value of x the corresponding vertical
line cuts the curve in more than one point, there will be
associated with such a value of x more than one value of y
(Fig. 12). The variable y is in such a case still a function of
X, since the values of y are determined by the values of x.
The distinction between functions of the latter type and those
previously considered is made by the following definitions :
If to every value of the variable under consideration there
corresponds a single value of the function, the function is said
to be singlevalued or onevalued. If to any value of the vari
able corresponds more than one value of the function, the
latter is said to be multiplevalued.
*The accuracy with which a function is defined by its grapb depends on
th6 accuracy with which it is possible to read the two scales of reference and
the ** fineness " of the curve.
I, § 20] REPRESENTATION OF FUNCTIONS
21
We shall for the present be concerned primarily with one
valued functions only, although one example of a twovalued
function will occur soon. Multiplevalued functions will be
considered later (Chapter X).
EXERCISES
1. From the following data construct a graph showing the cost of
domestic money orders in the United States :
Amount of Order
Kate
Amount of Okder
Katk,
•
Not over $ 2.50
3 cents
Over $30.00 to  40.00
15 cents
Over . 2.50 to 5.00
5 cents
Over 40.00 to 50 00
18 cent^
Over 5.00 to 10.00
8 cents
Over 50.00 to 60.00
20 cents
Over 10.00 to 20.00
10 cents
Over 60.00 to 75.00
25 cents
Over 20.00 to 30.00
12 cents
Over 75.00 to 100.00
30 cents
2. Draw a figure showing the rates for parcelpost packages for zone
1 ; for zone 2 ; for zone 3. Compare these graphs.
3. Draw a figure to represent the cost of gas in your own city. Is
there a different rate for large consumers ? If so, will this show clearly
on the graph ? How ?
4. On a piece of squareruled paper draw graphs of continuous func
tions which are rapidly increasing ; rapidly decreasing ; slowly increas
ing ; slowly decreasing.
5. Draw the graph of an arbitrary function which is increasing and in
which the rate at which it increases is increasing. Also that of an in
creasing function in which the rate of increase is decreasing.
20. Analytic Representation of Functions. We have
hitherto considered two methods of representing a function, the
graphic and the tabular. There is a third method, called the
analytic, which in its simplest form consists of the expression
of the function in terms of the variable by means of d, formula,
from which the corresponding values of the variable and the
function can be computed. The following will serve as examples.
22
MATHEMATICAL ANALYSIS
[I, § 21
21. Example 5. Capital and interest. The amount A in t years
of $ 1000 drawing simple interest of 5 % is given by the formula
(1) ^ = 1000 + 50<.
By substituting for t a suc
cession of values and com
puting the corresponding
values of A^ we obtain from
this formula a tabular rep
resentation of the function.
This in turn can be repre
sented graphically. The
a
1300
1250
1200
1150
1100
1050
1000
p
n
ri
n

1




■


"
■
"
~

'
"
'



■■


■
^
'
=:
'
^1
»
■

,
•f
■
'
/
<
.
i
V
6
Y
e
n
ra
t (years)
A (dollars)
1000
1050
1100
1150
1200
1250 1300
table above and Fig. 13 are the
result.* The points plotted ap
pear to be on a straight line.
Prove that they are.
22. Example 6. TJie
area of a square. The area
(in square inches) of a square
whose side is x inches long is
given by the formula
y = x2.
From this equation, we readily
compute the following table.
^
Jl
m t .
16 /
t
7
19 ~J
12 2
"F
5? y _ _ _
^ s t   ~
y
A_
, Z
^ 4
7
.<^
0^ 1 2 s 4 '0
Inches
Fig. 14
X (in.) . .
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
y (sq. in.).
0.25
1.00
2.25
4.00
6.25
9.00
12.25
10.00
* In practice bankers do not take account of fractions of a day in comput
ing interest. Strictly speaking, therefore, the graph of the function A, as
used in practice, is discontinuous. This practice of bankers is, however, dic
tated by convenience. It does not alter the fact that the function, as such, is
continuous.
1, § 24] REPRESENTATION OF FUNCTIONS
23
Using these values, it is now easy to draw the graph, which is shown
in Fig. 14.*
X
M.
*iL
Fig. 15
23. Example 7. The function de
fined by a circle. It is often desirable
to obtain an analytic representation of a
function, originally given graphically or
by means of a table. Such an analytic
representation is sometimes easy to obtain.
Suppose, for example, that on squareruled
paper an a:scale and a ?/scale have been
constructed with the units on the two
scales equals and suppose that with the
common 0point of the scales as a center
a circle is drawn with a radius of 2 units (Fig. 15). The functional rela
tion between the variables x and y defined by this curve is to be expressed
by means of a formula.
If P is any point on the circle, the absolute values of the x and the y of
this point form the legs "of a rightangled triangle of which the hypotenuse
measures 2 units. By a wellknown theorem of geometry we have then
y2 = 4 — a;2 or
y =± V4 — x^.
This is the analytic representation sought. It may be noted that we
have here to do with a twovalued function.
24. Range of a Variable. We had occasion some time ago
(§ 13) to introduce the term variable. In the future such a
quantity will generally be represented by a symbol, such as a;,
or 2/, or t, etc. Indeed this was done in some of the preceding
examples. The various values attached to such a symbol
throughout the discussion are numbers. These numbers con
stitute the range of the variable in question.
The range of a variable is usually determined by the nature
of the problem under consideration. Often it is very definitely
restricted. Thus in the case discussed in the last article the
* When, as here, the only fractional parts of a unit which occur are halves,
quarters, etc., it is convenient to use a ruled paper on which the larger units
are subdivided into four or eight parts instead of ten.
24 MATHEMATICAL ANALYSIS [I, § 24
range of x (as well as that of y) consists of all (real) numbers
from — 2 to + 2, and no others. For numbers outside this
range, the function in question is not defined. Again, in the
case of the mortality table considered in Ex. 5, p. 17, the range
of the dependent variable (the number of persons living at a
given age) is restricted to whole numbers less than 100,000 ;
fractional values of the variable are here meaningless.
25. Increasing and Decreasing Functions. A function
which increases when the variable increases is called an in
creasing function ; if, on the other hand, the function decreases
as the variable increases, the function is called decreasing.
Thus the amount A of capital and interest recently considered
is an increasing function of the time t, throughout the range of
the latter. Also, the area of a square is an increasing function
of the length of one of its sides. On the other hand, the num
ber of people livhig at a given age is a decreasing function of
the age. A function may be increasing for certain values of
the variable and decreasing for certain other values. Thus,
the temperature is during certain portions of the day an
increasing function, during other portions a decreasing func
tion. The volume considered in § 15 is a decreasing function
of the temperature T, from T = to T = 4, and an increasing
function for values of T greater than 4.*
If the two scales with reference to which the grapli of a function is
constructed are jjlaced in the more usual way, so that the numbers on the
scales increase to the right and upward, respectively, what distinguishes
the graph of an increasing function from that of a decreasing one ?
* In the case of the circle discussed in § 23, the function has two " branches "
in the interval from x = — 2 tox=42, the one consisting of the positive
values of y, the other of the negative values of y. The function may be con
sidered as consisting of two onevalued functions, one of which increases from
a;= — 2 to a; = and decreases from x = to x=\2, while the other de
creases from x=— 2 to a; = and increases from a; = to x=\2.
1, § 25] REPRESENTATION OF FUNCTIONS 25
EXERCISES
1. If a body falls from rest, its speed v in feet per second at the end
of t seconds is given by the relation v = 32 t. Construct the graph of v as
a function of t.
2. The charge for printing n hundred circulars of a certain kind is
) = 2 n + 10 dollars. Represent the function graphically.
3. The express rate r on a package is computed from the following
formula : r =^(p — 30)+ 30, where w is the weight of the package in
100
pounds and^ is the charge per hundred pounds. Draw the graph of r as
a function of w, for each of the values p = 40, 60, 80, 100. What com
ment would you make on this rule for p = 30, or for values of p less than
30 ? This is an example in which the range of the variable is arbitrarily
limited to be not less than a certain amount. The formula in this exer
cise really gives r as a function of the two variables w and p.
4. When a body is dropped from a height of 200 ft., its distance s
from the ground at the end of t sec. is given by s = 200 — 16.1 1^. Draw
the graph of s as a function of t. In how many seconds will the body
reach the ground ? At what time is the speed of the body greatest ?
Least ? What relation has the steepness of the graph to the speed of
the body ? Why ? What are tlie natural limitations on the range of the
variable ?
5. In Fig. 13, the beginning of the J.scale does not appear on the
graph. Why ?
6. Rate of increase. In the function of § 21, when < = 2, we have
A = 1100. Starling with this initial value of t, let f be increased by 1, by
2, by 3, ••• The corresponding values of A (i.e. the values of A when
^=2 + 1 = 3, 2 + 2=4, etc.) are respectively 1150, 1200, 1250, •■., and
the corresponding increases in A over the initial value 1100, are 50, 100,
150, •••. We see then that for these values the increase in ^ is always
equal to 50 times the corresponding increase in t.* Show that the same is
true if we start with a different initial value of f, say t = 3. Prove, in
general, that starting with any particular value, say t '= ti, of t, and any
increase in t, say an increase equal to h, that the resulting increase in A
is equal to 60h; i.e. that the ratio
increase in A ^q
corresponding increase in t
* When a change in the value of the variable produces a certain change in
the value of the function, these two changes correspond to each other. We
may then speak of either change as corresponding to the other.
26 MATHEMATICAL ANALYSIS [I, § 25
7. From the result of Ex. 6, show that the graph of the function there
considered is a straight line.
8. Make an investigation similar to that in Ex. 6 for the function y=x^
considered in § 22 ; i.e. calculate the increase in y due to an increase
in X, under a variety of conditions. For example, let x = 2 initially, and
calculate the increases in y resulting from increases of 0.5, 1.0, 1.5, 2.0 in
X. For each case calculate the ratio :
increase in y
corresponding increase in x
Is this ratio constant ? Is the increase in y due to an increase in x of 1.0
the same when the initial value of x is 3 as it is when the initial value of x
is 2 ? How is the change in the steepness of the graph related to your
result ?
9. A car begins to move and gradually increases its speed in such a
way that in x sec. it has traveled y = x^ ft. Interpret in this new setting
the "increase in y due to a certain increase in x," as computed in the
preceding exercise. Show in particular that the "increase in y" is the
distance traveled by the car during the interval of time represented by
the corresponding " increase in x," and that the ratio
increase in y
corresponding increase in x
is the average speed of the car during this interval. Does this suggest
a method for computing approximately the speed of the car at a given
instant ?
10. A certain function y has the value 0, when the variable x is 0, and
has the value 4, when x = 2. The graph of the function is a straight line.
Draw the graph and tabulate, from the graph, the values of y when x=l,
3, 4, 5, C. What is the algebraic relation between y and x ?
11. The graph of a certain function is a straight line. Draw this
graph, knowing that y = 0, when x = — 1, and that y = 4, when x = 3.
Discover the equation connecting y and x.
26. Statistical Graphs. One of the most generally familiar
uses of the graph is in connection with the representation of
statistical data. The figure below represents the enrolment in
Dartmouth College during the years 19051915. The method
of its construction should be clear without further ex
planation.
I, § 26] RPJPRESENTATION OF FUNCTIONS
27
An essential difference between this sort of graph and those
previously considered must, however, be noted. Strictly speak
ing, the graph consists only of the points forming the corners
\ U4^fr[f > HJ ^ H 1
^
1400 T  " "f
T :: z''/
:::::::::::::::::: ,,::„:^^: ::::::::
„ 1300 J r ± ±':^rk^ 
^ ::::::::::::::::::::::::::::ii"::L4::::::::: ::::::::::
S 1200 ;H; p^ 1  
CO ^ r**'''
"^ 1 1 1 1 1 1 1 1 li^44Rfem 1 1 1 1 1 1 ! 1 1 1 mm 1 1 1 1 1 1 1 1 1 nTrrl
^ 1100 2 "'" X "
1 ::::>^::::;^:^:._±
^ ^"^l \y(i[[\\\\\ III
900
1905 1906 1907 1908 1909 1910 1911 1912 1913 19li 1915 Years
Enrolment of Dartmouth College, 19051915
Fig. 16
of the broken line in the figure. The dates, 1905, 1906, ••• refer
to the beginning of the college year in September of the years
given, and the points plotted give the enrolment at the begin
ning of each such year. The straight lines joining these points
are drawn merely for convenience, as an aid to the eye in follow
ing the changes in the enrolment from year to year. The points
of these lines between the end points have no significance. The
range of the variable here consists of the finite number of dates,
1905, 1906, •••, 1915; and the function considered is discontin
uous. In such a graph interpolation is obviously impossible.
Questions
(1) During what periods did the enrolment increase ? decrease ?
(2) What was the percentage of increase during the 11 years ?
(3) What was the average rate of increase (in students per year) from
1905 to 1915?
28
MATHEMATICAL ANALYSIS
[I, § 26
(4) If the first point of the graph (1905) be joined to the last point
(1915) by a straight line (see figure), how is the steepness of this line re
lated to the average rate of increase ?
EXERCISES
1. The maximum temperatures (in degrees Fahrenheit) at Hanover,
N.H., on successive days from Oct. 1 to Oct. 15, 1914, were respectively as
follows :
59.6, 74.8, 79.7, 82.1, 78.9, 66.6, 61.4, 73.7, 82.5, 73.2, 78.9, 66.8, 55.0,
67.0, 63.5.
Construct a graph representing these data by a broken line. Is inter
polation possible ? Why ?
2. American shipping statistics give the total iron and steel tonnage
built in the U.S. for the years 19001914 as follows :
Year
Tonnage
Year
Tonnage
Year
Tonnage
1900
196,851
1905
182,640
1910
250,624
1901
202,699
1906
297,370
1911
201,973
1902
280,362
1907
348,555
1912
135,881
1903
258,219
1908
450,017
1913
201,665
1904
241,080
1909
136,923
1914
202,549
Draw the graph. Is interpolation possible ? Why ?
27. Summary. As has already been sufficiently indicated,
the object of our work thus far has been to make clear the con
cept of a function. To this end we have considered a variety
of special functions. Confining ourselves at present to the con
ception of what we have had occasion to define. as a single
valued function of one variable, we have seen that the essential
characteristic of such a function may be defined as follows :
A variable y is said to be a function of another variable x, if
when a value of x is given, the value of y is determined.
A variable is a quantity which throughout a given discussion
assumes a number of different values. The values which a
I, § 28] REPRESENTATION OF FUNCTIONS 29
variable may assume constitute the range of the variable in
question.
The range of a variable may be limited or not according to
circumstances.
We have become acquainted with three methods of repre
senting a function : the analytic, the tabular, and the graphic.
We have made a beginning in the classification of functions :
singlevalued and multiplevalued functions ; continuous and dis
continuous functions ; increasing and decreasing functions ;
functions of one variable and of more than one variable.
We have had occasion to note some of the questions that may
arise in the consideration of a function : To determine the
value of the function when the value of the variable is given ;
the converse problem, to determine the value (or values) of the
variable, corresponding to a given value of the function. Both
of these problems may involve the process of interpolation.
The maximum or minimum value of a function (and the
value of the variable for which this maximum or minimum
occurs) is often of importance. So also is the rate at which
a function changes its values. This, we have seen, is in
timately connected with the steepness of the graph of the
function.
28. Algebra as a Tool. The methods to be used in the
future for the study of functions and their applications group
themselves naturally under three headings corresponding to
the methods of representing a function : graphs, analysis,
tables.
The first of these we have already considered. It has the
advantage of presenting the variation of the function vividly
to the eye ; in this respect it is the superior of either the
tabular or the analytic method of representation. It lacks
30 MATHEMATICAL ANALYSIS [I, § 28
precision, however, since any graph drawn on a piece of paper
is in the nature of the case an approximation.*
The analytic representation by means of a formula we have
touched only very briefly. One of its chief advantages is that
of the utmost precision and conciseness. This very conciseness,
however, tends to obscure the properties of the function. The
tools which enable a sufficiently skillful operator to bring out
the hidden properties inherent in a formula are comprised in
what is known as mathematical analysis, of which the processes
of elementary algebra form the foundation.
The more important functions have been tabulated. Such
tables are used primarily to facilitate numerical computations.
We shall have occasion to use tables frequently.
The next chapter is devoted to a brief discussion of certain
algebraic processes and of their relation to the graphic rep
resentation already discussed.
QUESTIONS FOR REVIEW AND DISCUSSION
1. Give examples from your own experience of quantities that are
functionally related. In each case, state as many properties of the function
as you can (continuous or discontinuous, increasing or decreasing, etc.).
2. State some general laws and discuss the functional relations they
illustrate.
3. Would it be desirable to define a function as follows : y is a function
of x, if y changes its value whenever the value of x changes ? Why ?
4. Give, from your experience, concrete examples of the use of an
arithmetic scale. Of an algebraic scale. What are the distinguishing
characteristics of these two scales ?
5. Describe the three methods of representing a function and discuss
the advantages and disadvantages of each.
6. If the graph of a function y of x is a straight line, and the value of
the function is known for a; = 4 and for x = 5 (say these values are 20
and 26, respectively), how can the value of the function for aj = 4.5 be
calculated (not read from the graph) ? For x = 4.2 ? For a: = 6.7 ?
*0n the other hand, we can conceive, theoretically, of a graph which is en
tirely accurate.
I, § 28] REPRESENTATION OF FUNCTIONS
31
MISCELLANEOUS EXERCISES
1. The following table gives the pressure of wind in pounds per square
feet in terms of the velocity of the wind in miles per hour :
Miles per hour
5
10
15
20
30
40
50
60
70
80
Lb. per sq. ft.
0.1
0.5
1.1
2.0
4.4
7.9
12.3
17.7
24.1
31.5
Represent the function graphically. Determine approximately the
velocity which will produce a pressure of 10 lb. per square feet. What
does the increasing steepness of the curve signify ?
2. The following table, prepared by the U.S. "Weather Bureau, gives
the average monthly values of relative humidity at the stations given :
C3
<
5
><
•5
6
<
5
o
>
o
p
New York
75
74
71
68
72
72
74
75
76
74
75
74
Chicago . .
82
81
77
72
71
73
70
71
70
72
77
80
New Orleans .
79
80
77
75
73
77
78
79
77
74
79
79
San Francisco
80
78
78
78
79
80
84
86
81
79
77
80
Plot on the same sheet of paper. Is interpolation possible ? Why ?
3. The following table gives the average weight of men and women for
various heights :
Height .
Weight in Lb.
Height
Weight in Lb.
Men
Women
Men
Women
5 ft.
5 ft. 2 in.
5 ft. 4 in.
5 ft. 6 in.
128
131
138
145
115
125
135
143
5 ft. 8 in.
5 ft. 10 in.
6 ft.
6 ft. 2 in.
154
164
175
188
148
160
170
Represent the two sets of data on the same paper and draw any conclu
sions that seem reasonable. Is interpolation possible ? Why ?
4. The attendance at a base ball park on successive days was as follows :
1002, 1800, 1875, 1375, 1500, 2750, 3520. Represent these data by points
in a plane. Is a curve drawn through these points of any significance ?
Explain your answer.
32
MATHEMATICAL ANALYSIS
[I, § 28
5. The London Economist gives the following table showing the net
tonnage of steamships and sailing vessels on the register of Great Britain
and Ireland from 1840 to 1912 :
Year
Steamshii'
Sailing
Vessel
Year
Steamship
Sailing
Vessel
1840
1860
1880
1900
87,930
464,330
2,723,470
7,207,610
2,680,330
4,204,360
3,851,040
2,096,490
1909
1910
1911
1912
10,284,810
10,442,719
10,717,511
10,992,073
1,301,060
1,112,944
, 980,997
902,718
Represent these data graphically on the same sheet of paper. What
fact does this graph vividly portray ?
6. The temperature drop t below 212° at which water will boil at differ
ent elevations and the elevation h in feet above sea level are connected by
the relation h =1"^ + 517 1. Construct a table of values of h ior t = 0, 5,
10, 15, 20, 25, 30, and draw the graph of 7i as a function of t. At what
temperature will water boil on Pike's Peak, 14,000 feet above sea level ?
About how high is it necessary to go in order that water will boil at 200"^ ?
CHAPTER II
ALGEBRAIC PRINCIPLES AND THEIR CONNECTION
WITH GEOMETRY
29. Numbers and Measurement. We have already had
occasion to distinguish between two kinds of numbers :
(a) Numbers each of which represents a magnitude only ;
(6) Numbers each of which represents a magnitude and one
of two opposite senses, i.e. the socalled signed numbers.
It seems desirable at this point to recall the familiar classifi
cation of these numbers and the way in which they serve to
give the measures of magnitudes. We confine ourselves first
to the numbers of Type (a) above.
Integers. The first numbers used were the socalled whole
numbers or integers,
1, 2, 3, 4, ...,
which represent the results of counting and answer the ques
tion : How many ? They also represent the results of measure
ments, when the magnitudes measured are exact multiples of
the unit.
The Rational Numbers. When the magnitude measured
is not an exact multiple of the unit of measure, other num
bers called fractions must be used.
•' A\ 1 1 1 1 *B
These numbers are intimately asso c^ . 1 1 ,/>
ciated with the idea of a ratio. ^' ^^
Fig. 17
Thus, in geometry, two line seg
ments AB and CD are called commensurable, if there exists
a third segment PQ of which each of the other two is an
D 33
34
MATHEMATICAL ANALYSIS
[II, § 29
exact multiple (Fig. 17). PQ is then called a common measure
of AB and CD. If AB is exactly m times PQ and CZ> is
exactly n times PQ, m and n being integers, we say that the
ratio of AB to CD is m/n, and we write
AB^m
CD n'
If CD is the unit of length, we have
the measure ofAB — — •
A number which can be written as a fraction in which the
numerator and denominator are both integers is called a
rational number*
Such numbers suffice to represent the measure of any magni
tude which is commensurable with the unit of measure.
The Irrational Numbers. If two magnitudes have no
common measure, they are called incommensurable. Thus we
know from our study of geometry that
the diagonal of a square (Fig. 18) is not
commensurable with one of its sides.f
Hence, the length of the diagonal of a
square whose side is 1 unit cannot be
expressed exactly by any rational num
ber. To meet this deficiency the socalled
irrational numbers, such as the V2, were
introduced.
It is beyond the scope of this book to treat irrational num
bers fully. But we may note that they serve to express the
* Observe that according to this definition the rational numbers include
the integers. The number "zero" is also classed among the rational num
bers. See § 30.
t If AB and AC had a common measure /, such that AB = m x I and
AC = nX I, where ?n and n are integers, it would follow that n^ = 2in'^ ; but
this relation cannot hold for any integers in and «. Why ?
II, § 29] ALGEBRAIC PRINCIPLES 35
ratio of pairs of incommensurable magnitudes, and, in particular,
to express the measure of any magnitude which is incommensur
able with the unit.
Moreover, any irrational number may be rejjresented approxi
mately by a rational number with a7i error which is as
small as we please. This follows from the following con
siderations.
It is important to note that the result of any actual direct
measurement is always a rational number. For example, in
measuring a distance, we use a foot rule marked into fourths,
or eighths, or thirtyseconds of" an inch, or else some more
accurate instrument divided into hundredths or thousandths of
a unit, and we always observe how many of these divisions are
contained in the length to be measured. The result is, therefore,
always a rational number m/n where n represents the number
of parts into which the unit was divided. Any such actual meas
urement is, of course, an approximation. The greater the ac
curacy of the measurement (and this accuracy depends among
other things on the number of divisions of the unit) the closer is
the approximation. Since we may think of the unit as divided
into as many divisions as we please, we may conclude that any
magnitude can be expressed by a rational number to as high a de
gree of accuracy as may be desired. Thus, the length of the
diagonal of a square whose side measures 1 in. is expressed
approximately (in inches) by the following rational numbers :
1.4, 1.41, 1.414, 1.4142. These decimals are all rational ap
proximations, increasing in accuracy as the number of decimal
places increases, to the irrational number V2.*
* Surds, i.e. indicated roots of rational numbers, are not the only irrational
numbers. The familiar n = 3.14159 ••• is an example of an irrational number
which is not expressible by means of any combinatiou of radicals affecting
rational numbers.
36 MATHEMATICAL ANALYSIS [II, § 30
30. The Number System of Arithmetic. The (unsigned)
rational and irrational numbers, together with the number zero
(which is counted among the rational numbers), constitute the
number system of arithmetic.
31. The Nimiber System of Algebra. Corresponding to
any unsigned number a (except 0) there exist two signed
numbers + a and — a. The magnitude represented by a
signed number is called the absolute value of the number, and
is indicated by placing a vertical line on each side of the
number. Thus the absolute value of + 5 and of — 5 is 5 ; in
symbols, +5= — 5=5.
The signed numbers are called rational or irrational accord
ing as their absolute values are rational or irrational. The
entire system of positive and negative, rational and irrational,
numbers and zero * is called the 7'eal number system and any
number of this system is called a real number. These
numbers are contained in the socalled number system of
algebra.^
* Note that zero is neither positive nor negative. It has no sign.
t The number system of algebra contains also the socalled imaginary or
complex numbers, which will be discussed later. It may be noted that the
words rational, irrational, real, imaginary, are here used in a technical
sense. The popular meanings of the terms have no significance. V2 is no
more " irrational " (i.e. absurd or crazy) than the number 2 ; and the im
aginary numbers are just as "real" in the popular use of the term as are the
(technically) real numbers. Historically, the reason for the use of these
words is, however, connected with their customary meaning. For, while the
integers and rational numbers are of great antiquity, the irrational numbers
were not hitroduced until about the fifteenth century a.d., although incom
mensurable ratios were discussed by the ancient Greeks. At that time their
nature was not thoroughly understood, and it was not unnatural then to
designate them as irrational. Similar remarks could be made about the
introduction of the imaginary numbers toward the end of the eighteenth
century. We may add that what we now call "negative" numbers were in
the fifteenth century often referred to as " fictitious numbers."
II, § 32] ALGEBRAIC PRINCIPLES 37
32. Geometric Representation. Coordinates on a Line.
It follows from § 29 that the rational and irrational numbers
are just sufficient to express the length of any line segment.
Every segment on a line having one extremity at a given
point or origin can be represented by such a number;
and every such number will determine a definite one of these
segments, the unit of measure having been previously chosen.
This leads at once to the idea of an arithmetic scale, if we
confine ourselves to the numbers of arithmetic, and to the idea
of an algebraic scale, if we choose one of the directions on the
line to be positive, and use the real numbers of algebra to
represent the (now) directed segments. In the future we shall
generally confine our discussion to the algebraic case. No
confusion need arise from regarding an arithmetic scale as the
positive half of an algebraic scale, nor from regarding the
numbers of arithmetic as equivalent to the positive numbers
(and zero) of the real number system.*
It is often convenient to regard the number x which origi
nally represented the length and the direction from to a
P
Fig. li)
point P of the line as representing the point P itself, in which
case we call x the coordinate of P (Fig. 19). When we have
chosen a point as origin, selected a unit of length, and
specified which of the two directions on the line is positive, we
say that we have established a system of coordinates on the
line. When this has been done, every point P of the line is
represented by a number, and every real number represents a
definite point of the line.
* For this reason we shall often omit the + sign in writing a positive
number ; e.g. write simply 5 for + o. The context will always tell whether
the number in question is signed or not.
38
MATHEMATICAL ANALYSIS
[II, § 33
33. Coordinates in a Plane. We may now give the precise
mathematical formulation of the process already used (in con
nection with the construction of the graphs of functions) for
" plotting " points in a plane. The essential features of this
process are as follows (Fig. 20). We locate arbitrarily in the
^
Secona
quadrant
First quadrant
+ 5
,Pi
^
M.O
t/, ^
31,
X' 5
; " , J
X
quadrant
^s .5
■
Pi
Third
Fourth quadrant
Y'
Fig. 20
plane two algebraic scales, a horizontal one called the xaxiSj
and a vertical one called the yaxis. These two scales, called
the axes of reference^ intersect in the zero point of each scale ;
this point is called the origin. The position of any point P in
the plane is then completely determined if its distance and
direction from each of these axes is known. The units on the
two scales are arbitrary ; they may or may not be equal to
each other. The distance from either axis must, however, be
measured in terms of the unit of the other axis, i.e. of the axis
parallel to which the measurement takes place. Thus, in Fig.
20, where the units on the axes are different, the point Pj is
determined by the distance a; = 3 units from the yaxis (meas
II, §34] ALGEBRAIC PRINCIPLES 39
ured in terms of the ajunit) and the distance y = 2 units from
the a>axis (measured in terms of the y\mit). Similarly, the
points P2, P3, P4 are determined respectively by the directed
segments OM2 and M2P2, OM^ and M.^P^, OM^ and M^P^ ; the
numbers representing these directed segments are signed
numbers, so that the number gives both the magnitude and the
direction of the segment. In such a system of rectangular
coordinates in a plane, unless specifically agreed on otherwise,
the positive direction on the a;axis is always to the right; on
the 2/axis, always upward.
We see, then, that every point in the plane is determined
uniquely by a pair of numbers, and, conversely, that every
pair of (real) numbers determines uniquely a point in the
plane. The two numbers thus associated with any point in
the plane are called the coordinates of the point ; the number
X (giving the distance and direction from the ?/axi^) is called the
xcoordinate or the abscissa of the point, the number y (giving
the distance and direction from the ajaxis) is called the
ycobrdinate or the ordinate of the point. Any point P in
the plane may then be represented by a symbol {x, y), where the
abscissa of P is written first in the symbol and the ordinate of
P is written last. Thus we may write (Fig. 20) Pi =(3, 2)
P, = ( 1, 4), P3 = ( V2,  31), P, = (?, ?).
The two axes divide the plane into four regions called
quadrants, numbered as in the figure. The quadrant in which
a point lies is completely determined by the signs of the
coordinates of the point. Thus, the first quadrant is charac
terized by coordinates ({, +), the second quadrant by
( — , +), the third by ( — , — ), and the fourth by (+, — ).
34. Relations between Numbers. If two numbers a and
b represent two points A and B respectively on an algebraic
scale, we say that a is less than b (in symbols, a <b), it a is to
40 MATHEMATICAL ANALYSIS [II, § 34
the left of b, the scale being horizontal and the positive
direction being to the right.* The following obvious relations
are fundamental :
(1) If a =5t 6, then either a < &, or 6 < a.
(2) If a < 5 and 6<c, then a<c.
EXERCISES
1. Is the date 1916 a signed number ? (Does it represent simply a
duration of time or does it represent a time after some arbitrary fixed
time ?) Would it be proper to represent the year 50 a.d. by + 50 and
the year 50 b.c. by — 50 ?
2. When we designate the time of day as " two o'clock," is " two" a
signed number ?
3. Are the (unsigned) integers used for any other purposes than to
express the result of counting or measuring ? ( House numbers, catalog
numbers, •••)
4. State some theorems of geometry concerning ratios.
5. Find a rational approximation of VS accurate to within 0.001.
6. Why is any actual measurement necessarily an approximation ?
7. Why is it incorrect to define a rational number as one " which does
not contain radicals ? "
8. Why should irrational numbers be used at all, if it is possible to
represent any such number by a rational number to as high a degree of
approximation as may be desired ?
9. Explain /rom the definition of ratio why  in. and f^ in. represent
the same magnitude. Why m/n in. and pm/pn in. represent the same
magnitude,
10. Two segments measure  in, and f in., respectively. Show that
the ratio of the first to the second according to the definition is y^^. (Ob
serve that ^ in. is a common measure of the two segments.)
11. Two segments measure m/n and p/q in. respectively. Prove that
the ratio of the first to the second is mq/np. (Find a common measure
of the two segments.)
12. Given that a<6, can we conclude that a<6? Why?
Given that  a  >  6 1, can we conclude that a > & ? Why ?
* Likewise, a is greater than b (in symbols, a > b), if ^ is to the right of
B. Obviously, if a < 6, then 6 > a.
II, § 35j ALGEBRAIC PRINCIPLES 41
13. Which is the greater, 3 or 4? —Slor — tr?
14. Locate on a line the points whose coordinates are 2, — , , — 2,
6. What is the distance between the last two ? What signed number
represents the directed segment from the point + 5 to the point — 2 ?
15. Locate in a plane the points (2, 3), (— 2, 3), (2, — 3), (—2,3),
referred to a system of rectangular coordinates, the units on the two axes
bfiing equal.
16. If the abscissa of a point is positive and its ordinate is negative,
in what quadrant is the point ? If abscissa and ordinate are both
negative ?
17. If the abscissa of a point in a plane is + 2, where is the point ?
If the ordinate is zero ? What characterizes the coordinates of a point
on the yaxis ? On the xaxis ? What are the coordinates of the origin ?
18. The units on the two scales being equal, what is the distance of
the point (3, 4) from the origin ? Of the point (— 1, 7) ? Of the point
(2,  1) ? Of the point (a, b) ?
35. The Fundamental Operations. We shall now take up
briefly the fundamental operations of addition, multiplication,
subtraction, and division, and develop certain geometric inter
pretations and applications connected with these operations,
which are of importance in what follows.
Addition. We note first that the operation of addition for
signed numbers has an essentially different meaning from that
for unsigned numbers. The addition of two unsigned numbers
expresses simply the addition of magnitudes. Thus, any two
magnitudes may be represented geometrically by the lengths
of two line segments. The segment, whose length represents
their sum, is obtained by simply placing the segments end to
end to form a single segment. (Compare the process of graphic
addition described in Ex. 3, p. 7.)
A signed number, on the other hand, represents a direction
as well as a magnitude ; it is represented geometrically by a
directed segment. Consider two signed numbers a and b.
They will be represented by two directed segments whose
42 MATHEMATICAL ANALYSIS [II, § 35
lengths are  a and 1 5 1, respectively, and whose directions are the
same or opposite according as the numbers have the same or
opposite signs. Figure 21 represents the four possible cases.
The sum a + 6 is represented by a directed segment which
expresses the net result of moving in the direction represented
a
y 1
b
I
1
a + b
1
a\
1
1
b
'
!
'
i — ds:
a+F o,+b
Fig. 21
by a through a distance equal to  « , and then moving in the
direction of b through a distance equal to \b\. The segment
representing a f 6 is the segment from the initial point of
these motions to the terminal point. (See Fig. 21.)
The difference in the meaning of addition in the case of unsigned and
signed numbers is clearly brought out by considering a simple concrete
example : Suppose you walk to a place five miles distant and back again .
The total distance you have walked is 5 + 5 = 10 miles. These are un
signed numbers. On the other hand, if you represent the trip out by + 6
and the trip back by — 5, which is entirely proper, the sum (+5)f(— 5),
which is equal to 0, does not represent the distance walked at all, but does
represent the net result of your walk measured from your starting point.
The total distance walked is represented by  + 5  f  — 6 .
It should be noted that the absolute value of the sum of
two numbers is not, in general, equal to the sum of their
absolute values. In fact all we can say in general on this
point is that
(1) k + ?><« + ^.*
The equality sign holds only when a and h have the same sign.
* The symbol "^ is read " is equal to or less than."
II, § 35] ALGEBRAIC PRINCIPLES 43
The geometric interpretations on the algebraic scale of add
ing a number x to all the numbers of the scale consists of
sliding the whole scale to the right or left, according as x is
positive or negative, through a distance equal to  a;. Figure 22
illustrates this operation for the value x = — 2.
Every number in the upper scale is the result of adding — 2
4 3 2 1 +1 +2 +3
'4 32 1 + i +2+3 +4
Fig. 22
to the number below it in the lower scale. Two important
consequences follow from this interpretation :
(1) If a <C b and x is any (real) number, then a { x <. b \ x.
(2) If a point P ivhose coordinate on a line is x is moved on
the line through a distance and in a direction giveyi by the number
h, the coordinate x' of its new position is given by the relation
(2) x' = X + /i.
An immediate consequence of the meaning of addition in
the case of directed segments is as follows. If A, B, C are
any three points on a line, then
(3) AB \BC = AC,
This relation holds no matter what the order of the pomts
on the line may be. In fact it is obvious that to move on a
line from Ato B and then to move from B to (7 is equivalent to
moving directly from A to C, no matter how the points are
situated on the line. As a special case of this relation we
have
AB\BA = 0, or AB = BA.
44 MATHEMATICAL ANALYSIS [II, § 35
Multiplication. The product ab of two signed numbers
a and b is defined as follows :
(1) a* = a.16.
(2) The sign of ab is positive or negative according as the
signs of a and b are the same or opposite.
The statement (2) involves the familiar law of signs :
(+)(+)=(+), (+)()=()(+)=(). ()()=(+)•
Geometrically, multiplication by a positive number x is
equivalent to a uniform expansion or contraction of the scale
away from or toward the origin in the ratio  «  : 1, according
as I a; I is greater than or less than 1.
This statement will become clear on inspection of the follow
ing figure (Fig. 23) which gives the construction for the multi
FiG. 23
plication of every number on the scale by x. In the first figure
X has been taken equal to + 2, in the second equal to \ ^.
The geometric interpretation of multiplication by a negative
number x consist? of a similar expansion or co7itraction in the
ratio I ic 1 : 1 combined with a rotation of the whole scale about
the origin through an angle of 180°. For such a rotation
will change each positive number into the corresponding nega
tive number, and vice versa, which the law of signs requires.
Here again we may note two consequences of importance :
1. If a < b and x is any (real) number, ax is less than, equal
to, or greater than bx, according as x is positive, zero, or negative.
2. If a scale is uniformly stretched (or contrax^ted), the origin
II, § 35] ALGEBRAIC PRINCIPLES 45
remaining fixed, in sugJi a way that the point 1 moves to the point
whose coordinate is a, then the point whose coordinate is x will
mx)ve to the point whose coordinate is
(4) X' = ax.
Subtraction. To subtract a number b from a number a
means to find a number x such that x{b = a. We then write
x = a — b.
Such a number x can always be found. Representing a and
b by directed segments having the same initial point, the
meaning of addition tells us at once that
the segment from the terminal point of < — 
b to the terminal point of a represents
the number x sought. (See Fig. 24.) ^'° ^
This shows, moreover, that to subtract a number b is equivalent
to adding the number — 5.*
Division. To divide a number a by a number b means to
find a number x such that bx = a. We then write x = a/b.
It is always possible to find such a number x, except when the
divisor b is zero. For we need merely reverse the construction
given for multiplication (Fig. 23) as
indicated in Fig. 25, first drawing
the line joining b on the original
scale to the point a on the multi
i 6\ plied scale and locating the required
Fig. 25 point x on the multiplied scale by
a line through 1 on the original scale, parallel to the line ab.
In particular, we can always find a number x such tha!
* It may be of interest to recall here the fact that historically the negative
numbers were introduced in order to make the operation of subtraction
always possible (i.e. even in the case when the subtrahend is greater than
the minuend). But from what has just beeu said it appears that the device
adopted for rendering the operation of subtraction more useful and convenient
had the additional effect of making this operation unnecessary.
46
MATHEMATICAL ANALYSIS
[II, § 35
bx = l,ifb=^ 0. This number 1/6 is called the reciprocal of h.
Hence, to divide by b (b =^ 0) is equivalent to multiplying by 1/6.
The Case 6 = 0. This case demands careful attention. Since
• « = for every number x, it follows that the relation • x = a
cannot be satisfied by any value of x, unless a is also ; and ivill
be satisfied by every value of x, if a is 0. Hence, by the definition
of division, the indicated quotient
x^
has no meaning whatever when a ^0, and no definite mean
ing even when a = 0. Hence, we conclude that division by
zero, being either impossible or useless, is excluded from the
legitimate operations of arithmetic and algebra.
36. The Function a/x. The Symbol oo . Whereas we have
just seen that division by zero is not a legitimate operation, it
is highly important for us to note what happens to the fraction
a/x when x assumes values approachiyig nearer and nearer to
zero ; as long as x does not equal zero, the indicated division
is possible. We wish then to consider the /;mcfio?i a/x = y for
values of x near 0. A table of corresponding values of x and y
is as follows :
X
4
3
2
1
\
\
4
3
2
 1
I
I
:
ia
i«
ha
a
2a
4.a
\a
i«
la
— a
2a
4a
Plotting the points (x, a/x) with reference to two rectangular
axes we obtain Fig. 26, where we have assumed a to be posi
tive and have chosen the unit on the ojaxis to be a times the
unit on the 2/axis.
An inspection of the table and the graph shows us that as x
decreases in absolute value, a/x increases in absolute value;
II, § 36]
ALGEBRAIC PRINCIPLES
47
more precisely, by choosing x sufficiently small in absolute
value, a/x can be made as large in absolute value as we please.
Further, when x = the expression a/x has no meaning.
We say the fmicUon is not defined for the value x = 0; or, the
range of the variable of this function does not include the value
X
Fig. 26
The sentence expressed in blackfaced italics above is some
times written in a species of shorthand :
00.
This looks like an equality involving a division by 0. But
it does not mean any such thing. The expression a/0 as indi
cating a division by has already been pronounced illegiti
mate. For this very reason we are at liberty to use the
symbol to mean something else without danger of confusion.
We accordingly use it as a short way of expressing the values
of the variable a/x as x is supposed to approach 0. Similarly,
48 MATHEMATICAL ANALYSIS [II, § 36
the symbol oo, read "infinity," does not represent a number at
all, but a variable which increases loithout limit. The above
equality is, therefore, an equality between variables, and is
simply a short way of writing the phrase " as the denomina
tor of a fraction, whose numerator is constant and different
from zero, approaches zero, the value of the fraction increases
without limit in absolute value." Under these circumstances,
we also say " the fraction becomes infinite." The phrase
" equals infinity," which is sometimes heard, is very mislead
ing and its use should be strictly avoided.
Returning to our table and graph, we note also that by
assigning to x a value sufficiently large in absolute value, the
value of a/x can be made in absolute value as small as we please,
but not zero. The shorthand expression of this fact is
00
or " as the denominator becomes infinite the fraction ap
proaches 0."
37. The Directed Segment P^Pz As an application of the
foregoing principles we will now derive a formula which will
often be used in the future. Let Pi and P^ be any two points
on an algebraic scale, and let their
coordinates be x^ and X2, respectively.
We desire to find the number repre
senting the directed segment P1P2
in direction and magnitude. By definition x^ = OPi, Xo = OP2
(Fig. 27). Now, by § 35, Eq. 3, we have
P1P2 = P,0 4 OP2 =  OPi + OP2
= Xi + X2,
or, finally,
/^i* 2 ^ •^2 — ^1*
Thus, if Xi = 2 and X2 = 5, X2 — x^ = \ 3, and we conclude
P2
P,
Xi Q Xi
Fig. 27
II, § 38] ALGEBRAIC PRINCIPLES 49
that the length of the segment P1P2 is 3 units and that its
direction is positive (i.e. from left to right in the ordinary
setting). On the other hand, if cci = 3 and Xg = — 4, we have
X2— X1 — —I, and we conclude that the length of the segment
is 7 and its direction is negative (i.e. P2 is to the left of Pj).
38. Concrete Illustration of the Law of Signs. The law of
signs, as indeed many of the fundamental laws of algebra, is essentially
a definition, arbitrary from a logical point of view and dictated largely
on the grounds of convenience. The following concrete example will
show how in one instance the conventions adopted in the law of signs
for multiplication correspond to the concrete facts to be described.
If a train moves at a constant speed of v miles per hour, then in t
hours it will travel a distance = ^?i miles. Here v, t, s are unsigned num
bers. Now, let us change the formulation somewhat, so as to introduce
the direction. At a given instant let the train be at a certain station 0;
let us count time from this instant (t = 0) so that any positive t desig
nates an instant a certain number of hours after the instant t = 0,
and a negative t designates an instant a certain number of hours before
t = 0. Further, let the position of the train be determined by the signed
number s representing the distance and the ^^
direction of the train from 0, s being positive [ ^ ^
if the train is to the right of (Fig. 28). =o *"■*"
Finally, let the speed and the direction in p^^ 28
which the train is moving be given by the
signed number v, v being positive if the train is moving to the right
(u = — 30, for example, would mean that the train is moving to the left
at the rate of 30 miles per hour).
Now consider the four cases : (1) v and t both positive ; (2) v positive
and t negative ; (3) v negative and t positive ; (4) v and t both negative.
Verify that the law of signs in the relation s = vt gives the sign to s
for which the actual position of the train in each case calls. [For
example : (1) If v and t are both positive, s = vt will be positive,
which is as it should be. For if the train is moving to the right, then
a certain number of hours after t = 0, when the train was at s = 0, it
will be a certain number of miles to the right of 0. (2) If v is positive
and t negative, s = vt is negative. This also is correct. For a train
moving to the right and arriving at O when t = 0, was to the left of
at any time before t = 0. Etc. ]
50 MATHEMATICAL ANALYSIS [II, § 38
EXERCISES
1. Under what conditions isa + & = \a\ +\b\?
2. Prove that if A, B, C, D, •••, i, M are any points on a line (in
any order) then AB + BC + CD + •" + LM = AM.
3. Graphic Addition. Given the directed segments, a, 6, c, d, e on
parallel lines (or on the same line), their sum a+6+c+cZ + e may be
found graphically as follows: On the straight edge of a piece of paper
mark a point ; lay the strip along the segment «, the point coincid
^  ing with the initial point of a ; mark the ter
'' *■ minal point of a on the paper. Then slide the
paper parallel to itself so as to make it lie
along h and bring the mark just made into
coincidence with the initial point of h ; mark
the endpoint of h. Then proceed similarly
for the segments c, d, e. The directed seg
ment from to the final mark will then represent the sum sought. Why ?
4. Draw directed segments representing the numbers — 3, + 5, + 2,
— 6, and find their sum graphically.
5. Show how to construct a directed segment representing the prod
uct of the numbers represented by segments a and h.
[Hint. Use the adjoined figure to determine the
magnitude of the product ; then determine the direc
tion. Observe that for the construction of a product
we need to know the length of the unit segment, which
was not necessary for a sum.]
6. Show how to construct a segment representing a/h.
7. Determine the numbers representing the directed segments from
the first point of each of the following pairs of points to the second: f 8
and + 6, + 8 and 6,2 and  4, — ^ and + , + 1.4 and  2.1, — 
and  I, + V and + 3.14.
8. By computing the numbers representing the segments, verify the
relation AB + BC = AC, when the coordinates of A, B, C are, respec
tively :
(a) 2, 3, 4 ; (6) 2,  3, 4; (c)  2, 3,  4 ; (d) 2,3, 4.
9. Find the coordinate of the midpoint of the segment joining the
points whose coordinates on a line are 4 and 8 ; — 3 and 5; — 2 and
— 5 ; Xi and X2.
II, § 40] ALGEBRAIC PRINCIPLES 51
39. Insight and Technique. Most of our activities involve
two more or less distinct aspects : insight and technique. On
the one hand, we need to understand the nature of the thing
we are trying to do, on the other we need skill in doing it.
Theory and practice, planning and carrying out the plans, etc.,
are other ways of pointing the same distinction.
In your previous study of arithmetic and algebra the major
emphasis was on the side of technique. You learned at that
time how to carry out certain manipulations with numbers ; and
you gained more or less skill in using the processes. In the
present course, the emphasis is to be placed on the side of in
sight, understanding, appreciation ; the technique of algebra is
to be used merely as a tool, not as an end in itself.*
40. Definitions. We propose now to recall very briefly a
few of the more important conceptions and processes of
algebraic technique. We shall begin with the definitions of a
few terms.
When two or more numbers are added to form a surriy each
of the numbers is called a term of the sum.
When two or more numbers are multiplied to form a 2)roduct
each of the numbers is called a factor of the product.
Any combination of figures, letters, and other symbols,
which represents a number, is called an expression. If the
equality sign (=) is placed between two expressions, the result
is called an equality, and the two expressions are called the
members or the sides of the equality.
An equality states that the two expressions represent the
same number. f
* However, we must maintain a certain amount of proficiency in the use of
algebraic processes. Hence " drill exercises " will not be wholly lacking in
what follows.
t Such a statement may or may not be a true statement. See § 47.
52 MATHEMATICAL ANALYSIS [II, § 40
Thus, suppose a, b, c, d, p, x, y represent numbers. Then
a — hx { 1 cdy = a (12 y"^ — p)
is an equality. The lefthand member is a sum of three terms;
the righthand member consists of only one term, which is
a product of two factors. The second term of the lefthand side
is a product of two factors, while the second factor of the right
hand side is a sum of two terms.
41. General Laws of Addition and Multiplication. The
following general laws we take for granted :
I. Concerning Addition :
1. Ariy tivo numbers may be added and their sum is a definite
member.
2. The terms of any sum may be rearranged and grouped in
any way without changing the sum.
Thus, if a, &, c, p, q represent any numbers whatever, we
have, for example, a {{b + c ^p) \ q ={b { q) \ {c \ a) \ p.
II. Concerning Multiplication :
1. Any two numbers may be nudtiplied and their product is a
definite number.
2. The factors of any product may be rearraiujed and grouped
in any way without changing the product.
Thus, if a, 6, c, x, y represent any numbers whatever, we
have, for example, {abc){axy)= a^bcxy ={yx){cba'^).
III. The Distributive Law : To multiply any sum by any
number m, we may multiply each term of the sum by m and add
the resulting j^foducts.
Thus, (a { b \ cd \ — \ x)m = am } bm f cdm + — + xm.
lY. The Law of Factoring : If every term of a sum con
tains the same number m as a factor , the sum contains m as a
factor.
Thus am \bm ^ cdm f ••• f xm = m (a f & + cd + — f x).
11. § 43]
ALGEBRAIC PRINCIPLES
53
Observe that IV is obtained from III by simply interchang
ing the sides of the equality.
42. Raising to Powers. Integral Exponents. We recall
also at this point the meaning and use of integral exponents.
The symbol cc", where x represents any number and n is any
positive integer, is an abbreviation for the product of n factors
each equal to x, i.e.
x"" = x X'X" to n factors.
From this definition and Principle II (§ 41) it follows at
once that
/pm^n =(x'X' X '" to m factors) (x • x 'X ••• to n factors)
= X ' X • X "• to m + n factors,
and therefore
Ya x»^x^ = x^^'\
Similarly
V6
— = x^^'^, if m > n,
Xn
Xn ;t"
if n > m.
(xmy = x"" • x"" ' x"^ •" to n factors
r^m+m+m+ ■•■ ton term"
and
Also
and therefore
VI
Also
Vila
VII6
43. Axioms. Closely connected with Principles I, 1 and
II, 1 are the familiar axioms
VIII a If a = h and c = d, then a ^ c = b ^ d.
VIII b Ifa = b and g = d, then ac = bd.
(^m)n
= x**'^^.
= a»b*K
64 MATHEMATICAL ANALYSIS [II, § 43
EXERCISES
1. Distinguish between insight and technique in the various professions
(surgery, dentistry, engineering, etc.).
2. Complete the following propositions :
(a) The sum of any two integers is • • •
(6) The product of any two integers is • • •
3. What is the familiar expression in words for Principle VIII ?
4. Find the results of the following indicated operations :
(1) a;i0a;i2. (6) x^^^x^ (11) (av^y.
(2) a^a^. 0) a^^a^^. (12) ( cM»y.
(3) ft^fes. (8) ^. (13) (0*.
a"
(4) y2ny3«. ^gj ^^4^8. (14) (r^S)".
(5) X'»%2. (10) (C2)6. ' (15) (X2)».
5. Multiply x^* + x^y^ + i/2& by x2» — x'^y^ + y"^.
6. Divide x^" + y^n by x" + ?/".
7. Perform the following operations :
(1)25.24= (2)25.44= (3)32.23= (4)7i5^7i3 =
44. Discussion of Principles. In the preceding article
Principles VVII were derived from I and II, while IV is a
consequence of III. We might now ask : " How do we know
that Principles I, II, and III are true for all numbers ? "
On these three principles the whole subject of algebraic
technique rests. They are so simple that they may appear at
first sight to be trivial. As a matter of fact their truth is by
no means obvious ; our unquestioned belief in them is the re
sult of experience in using numbers. Were we to attempt a
general proof, we should find it a long and difficult process
which is out of place in an introductory course. Hence we
simply take them for granted.
A little reflection will show that these principles are not obvious. Take
for example the fact implied by Principle II, 2 : a times b is equal to h
times a ; and let us suppose that a and b are positive integers. Now,
2x3 means 3 + 3 and 3x2 means 2 + 2 + 2. By addition we observe
II, § 44]
ALGEBRAIC PRINCIPLES
55
that the result is in both cases 6. But that simply verifies the general
law when a = 2 and 6 = 3. We can thus verify the law in question for
any two special integral values of a and h. Not only would this be ex
tremely laborious for large values ; it would still be only a verification for
a special case ; it would not be a general proof. Moreover, we have con
fined ourselves to the simplest of all numbers, the positive integers ; while
II, 2 asserts among other things that ah = 6a, no matter what numbers
a and 6 represent (rational or irrational, real or imaginary) . As has been
indicated in the preceding paragraph, we are not concerned in this course
with proving these principles. Of great interest to us, however, are the
relations existing between numbers and geometry. Accordingly we have
suggested in the exercises below some geometrical interpretations of these
principles. which furnish intuitive proofs of certain restricted cases.
EXERCISES
1. An intuitive proof that ah = ha, in case a and 6 are positive inte
gers : Let the integer a be represented by a group of a equal squares
placed side by side so as to form a row (see the figure, where a = 8).
The product 6 • a is then represented by the number of squares in 6 such
rows. Show that, if these rows be placed under each other (as in the
figure, where 6 = 6), it is seen that the total number of squares is also
equal to the number of squares in a columns each containing 6 squares.
Observe that while the figure is drawn for special values of a and 6, the
argument is general.
2. From a consideration of the adjacent figure
give an intuitive proof that 5 • (3 • 4) = 3 • (5 • 4) .
Then by using the fact that ah = ha show that
(3 • 4) • 5 = 3 • (4 . 5). Can this argument be made
general to show (a • 6) • c = a • (6 • c), when a, 6, c are positive integers ?
4
4
A
4
4
4
4
4
4
4
4
4
.4
4
4
66
MATHEMATICAL ANALYSIS
[II, § 44
3. From the adjacent figure, show how to use the idea of a rectangu
lar pile of blocks to prove that (a • 6) • c = a • (& • c), when a, &, c are
positive integers.
^^^4^^
7^
^
•^
j^'O^O^O' >'
^
u^
4. Assuming that the area of a rectangle is equal to the product of its
base by its altitude, show that ab — ba, when a, b are any positive real
numbers.
5. By considering the adjacent figure, interpret
a b
geometrically the relation {^a ■\ b)c := ac {■ be.
6. Interpret geometrically the equalities
(a) (a + &)2 = a2 + 2 a6 + &2.
(6) (a + 6) (c + (?)=: ac + he ^ ad ^ bd.
d
€
7. Derive the equalities in Ex. 6 from Principles IV.
[For 6 (6), we first consider c + d as a single number. Ill then gives
(a + 6)(c + d) = a(c + d) + 6 (c + d). Applying II and III to each of
the terms of the righthand member of this equality, we get the desired
result. ]
8. Show in detail how the carrying out of the product (ax+6)(cx+d)
involves Principles IV.
9. Show how these principles apply in the addition of 2 x^ + 7, 3 x + 2,
and 4 x2 + X } 3.
II, § 45] ALGEBRAIC PRINCIPLES 57
45. Review of Algebraic Technique. We propose now to
take up a few of the most elementary portions of the tech
nique of algebra. These are all that will be needed in the
immediate future. Other topics relating to technique will
be recalled when they are needed.
The technique of algebra is concerned altogether with the
transforming of expressions into other equivalent expressions
which serve better the purpose in hand. The principal pro
cesses used are the following :
(a) Performing indicated operations and collecting terms. For ex^^
ample, collect the terms in x, ?/, and z in the following : ^
2x472/ 3^; + y \x—^y ^bz^ 3x. T"..''/'
The result \%x \1z. This involves Principles I and IV.
Perform the indicated operation and collect terms in
(x2_3a;4)(x2).
The result is a;^ — 5 a;^ + 10 a: — 8. This involves Principles IV.
(b) The use of special products. The following equalities should be
memorized :
(1) (a}b)(a6)=a262.
(2) (a + b)2 = a^ + 2a& + &2.
(3) (a&)2 = a22ab + 62.
(4) (AC + a)(x h b)= x2 + (a + 6) X + ab.
(c) Factoring. The following cases may be specially mentioned :
i. I%e difference of two squares. Use special product (1). Thus
^9x^  iy^ ={7 x« + 2y)(7 x^  2y).
ii. Trinomials of the form x^ + px + q.
Try to find two numbers whose sum is p and whose product is q, in
accordance with special product (4) . Thus to factor
x2 _ 6 X  27
we notice that 3 and — 9 are two numbers which satisfy the requirements.
Hence,
a;2_6x27 = (x+ 3)(x9).
Again, to factor x2 — 10 x + 25, we notice that — 5 and — 5 are two
numbers satisfying the required conditions. Hence
x2 _ lOx + 25 = (x  5) (x  5) = (X  5)2.
58 MATHEMATICAL ANALYSIS [II, § 45
EXERCISES
1. Perform the following operations :
(a) (a; 4) (2 a: + 3).
(&) (x + a)(x + b)(x + c)'
(c) (x + h)^x^.
(d) (m + w) n — (m — n)m.
(e) (a + &)2(a6)2.
(/) (a + 6)3 (a 6)3.
2. Factor:
(a) x^W.
(&) (2aby9(xiy^.
(c) ax — bx + ay — by.
(d) x* 6x^ + 9.
(e) x2 + 6x + 6.
(;i) 25x* + 10a;3 + x^.
3. Factor:
(a) 4a2_5« + i.
(6) a2 + 2a6 + 62_a;2.
(c) a964a3_a6 + 64.
((^) 1 + x2 + xK
(e) x3 3a;2 + 2a;.
(/;2 + 7x15x2.
(Sr) x8 + 1.
(/i) a:4?/2 _ 17 x'^y  110.
46. Operations with Fractions. These depend chiefly on
the simple principle that the numerator and the denominator
of a fraction may be multiplied by the same number (not zero)
without changing the value of the fraction, and the reverse of
this principle, viz. that any common factor (not zero) of the
numerator and the denominator of a fraction may be removed
without changing the value of the fraction.*
By means of this rule any two or more fractions can be re
duced to the same denominator. The rules for adding and
multiplying fractional expressions are stated symbolically as
follows :
a
b
+r
b
c
a
b
xr
ac
'bd
a
b~
na
''nb
* This principle is a direct consequence of the definition of division. Can
you explain it ?
II, § 46] ALGEBRAIC PRINCIPLES 59
IBR^
iK
: PRINCIPLES
a
b
+r
ad\ be
bd
a
b_
a
"b
d
• — —
c
ad
= — .
be
d
a
b
= 
— a
b
a
b
a
+ &_
ab
Also
and
c c
The following exercises will furnish applications of these
principles.
EXERCISES
1. Express as a single fraction :
^^^ M^ + ^T^ + ^T&' ^ ^ c^y a'^ b^^
^ X y z x + y \z \n vj\m uj
(c) L^l + L. (^) 1 + 1_§. ,
St tr rs ^ 2 X y
Simplify the following expressions, assuming that no canceled factors
have the value zero.
a\b
2.
X 
— +
 a
{X
aY
\x
ay
3.
a 
 c
h + c
(ab){cb) {ac)(ab) {bc){ca)
Ans. 2«
(a_6)(c6)
«J=L^ _ 1±A I (62 _ an Ans. 4 ab.
a + b a&P ^
g x^ y* ^ gi^ ab + b^
' \a^ b^\ ' \a^ b^
60 MATHEMATICAL ANALYSIS [II, § 46
be ac ab a) ( i _ ^ c
V a + & + c
\a^^b^)\ b I \b a)
^2 ^. 7y2 q;2 _ ft2
)
10.
11.
aP  b^
a' + b^
a + b
ab
1
ab
a + b
^_^x^ 2\xy x\y)
1+^+ . "".J M
xy + 4 y'^ x^ + 4:xy y(x + 2 y)
Aiis. a
111 c U (a + 6)^c° l
I a + 6l (a + 6/ j
i + l + i
13. 2_L_2.
£ + « + *
a & c
14 a^ + &'^ ^ ■ ?) \ g^fe^
a^ + rt6 + ^n a bl a^ab + b^'
X^ + J^\ {X'^ + y^}
X'^ 1/2 '
15. ^ ^^ Ans. X*  a;2y2 + y*,
^ I y
xhy x — y
16. If a, ft, and « are positive, which is the greater,
or
b (b{x)
Distinguish two cases a> b^ a <.b.
17. If ad < eft, then is it true for all values of the letters involved that
a/b < c/d ? Why ?
II, §48] ALGEBRAIC PRINCIPLES 61
47. Identities and Equations. We must recall here a vital
distinction between two kinds of equalities. An equality
which is true for all values of the letters (or other symbols)
involved, for which both members of the equality have a mean
ing, is called an unconditional equality or an identity. An
equality which may be true for certain values of the letters
involved, but is not true for all, is called a conditional equality
or an equation. For example, the equality a'^ — b^=(a— b) (a + b)
is an identity since the two members of the equality represent
the same number for all values of a and b. Also the equality
^3^ =^ + ^
is an identity, even though it becomes meaningless when a = b.
Why ? On the other hand 2^ — 8 = is an equation since
it is true only for x = 4. Va? + 1 = — 1 is also an equation, but
it is not true for any value of x. Why ? *
To solve an equation is to find the values of the letters for
which it is true. Thus in the first example above, a; = 4 is the
solution or root of the equation 2 a? — 8 = 0. The second equa
tion above has no root.
48. The Relation AB = 0. In the solution of equations
the following principle is of frequent application. If a prod
uct of expressions each representing a number is zero, we may
conclude that some one of the factors is zero. In the simplest
case this means that if A and B represent expressions and if
A • B = Oy we conclude that either ^1 = or J5 = O.f
* By Va is meant the positive square root of a.
t We must, of course, be careful to assure ourselves that each of the expres
sions involved represents a number for the values under consideration. Thus
we cannot conclude from the relation x • (l/x) = 0, that either a = or 1/x = 0,
for when a; = 0, l/x is meaningless. In fact the given relation is impossible;
the equality is not true for any value of x.
62 MATHEMATICAL ANALYSIS [II, § 48
We may apply this principle to show the absurdity of some
mistakes that are often made by the careless student. For
example, a favorite mistake is to " cancel " the x in the expres
sion
a\x
This would be justified if the equality
a\ a + x ^a
^^ b+x b
were an identity. If we clear this equality of fractions by
multiplying both members by (b + x) b, we obtain
ba '\ bx = ab { ax,
or
bx = ax;
or, finally,
(ba)x = 0.
Hence we conclude that equality (4) cannot be true, unless
either 6 = a, or ic = 0. The " canceling operation " mentioned
above is therefore unjustified.
EXERCISES
1. Treat similarly the following equalities to determine under what
conditions they are true. Each one is related to an error that is some
times made.
(a) Is Va2 + 6'^ = a + 6 ? (Square both sides.)
^ ^ b d b\d
^ ^ 2c + d c + d
(d) Is (x + 2/)2 = x2 + y2 ?
(e) Given x^ = 2x. Are we justified in concluding that x = 2 ?
II, § 48] ALGEBRAIC PRINCIPLES 63
2. In each of the following equalities, assuming that the letters repre
sent real numbers, determine which are identities and which are equations.
Among the latter, distinguish those that are not true for any (real) values
of the letters involved ; and for the others determine in their simplest
form the conditions which they imply on the letters involved.
(a) a;*y4=(x + 2/)(xy)(x2+/2).
(6) x'^Sx + 2 = 0.
(c) x+ = 0.
X
(d) ac — he + ad — bd = 0.
3. Find and discuss the error in the following reasoning :
Let X = 2. Then a;2 = 2 x, and x^ — 4 = 2x — 4. This is equivalent to
(a;+2)(x2) =2(:b2).
Dividing both sides by x — 2, we get
cc + 2 = 2.
But X = 2 ; hence
2 + 2 = 2
or
4 = 2.
4. Find and discuss the error in the following reasoning : Let a and b
represent two numbers. Then
a2 _ 2 a6 + 62 =62 _ 2 a& + a%
or
(a.6)2=(6_a)2,
or
a — b = b— a;
hence
a = 6.
PART II. ELEMENTARY FUNCTIONS
CHAPTER III
THE LINEAR FUNCTION. THE STRAIGHT LINE
49. A Linear Function. Distance traversed at uniform speed.
Example. A railroad train starts 10 miles east of Buffalo
and travels east at the rate of 30 miles per hour. How far
from Buffalo is the train at the end of x hours ?
In oc hours the train travels 30 x miles. If its distance from
Buffalo is denoted by y, we have 2/ = 30 a; + 10. Pairs of values
of X and y obtained from this equation are shown in the fol
lowing table.
X
1
2
3
4
etc.
y
10
40
70
100
130
etc.
Geometric Eepresentation. Let us plot as points in a
plane these corresponding values of x and y. We then obtain
the first of the following figures (Fig. 29). It will be noticed
Y
Y
Y
I '
• 3 4X i i
^ ^X 12 3 4
Fig. 29
64
Ill, §49]
LINEAR FUNCTIONS
65
that the five points appear to lie on a straight line. We have,
for intermediate values of x, the values of y shown in the fol
lowing table.
X
i
^
■ 2i
H
y
. 25
55
85
115
Plotting these points we obtain the second of the above figures,
in which the nine points appear to lie on a straight line. Let
us calculate the value of y for some more intermediate values
of X thus :
X
JL
4
f
i
i
y
m
?>2i
47i
62^
In the third figure we see that these new points still appear to
lie on the same straight line.
These considerations suggest that if we could calculate the
values of y corresponding to all the values of x between x =
and a? = 4, the points whose coordinates
are {x, y) would all lie on a straight
line joining the points (0, 10) and
(4, 130), and would constitute the
whole of this linesegment. A proof
that this is the case is as follows : In
Fig. 30 we have drawn the straight
line joining the points A (0, 10) and
B (4, 130). Let (x^, y^)(x, > 0) be any
pair of corresponding values of x and y for the function
2/ = 30 ic + 10 ; we then have
(1) 2/1 = 30 0^1+10.
. ,,,
120 ^
 J «!
i^
80  Al
60' :::^ :n
^'^ Jz t^
40  ^ ^ 1 ^
/^ II
20 ^ til '
A 4  yj, ^
^ iiSi^SElL
1 2 M^ *
Fig. 30
66 MATHEMATICAL ANALYSIS [III, § 49
We now wish to prove that the point P (a^i, y{) is on this
line AB.* To do this we construct the triangles APQ and
ABC by drawing lines through A, P, and B parallel to the
axes. If P is on the line JLB, then these triangles are similar,
and if P is not on the line AB, then the triangles are not similar.
Why? If the triangles are similar, QP/CB = AQ/AC; and
conversely, if QP/CB = AQ/AC, the two triangles are similar.
Expressed in terms of the coordinates of A, By and P, this pro
portion becomes (see figure)
120 4 '
or ^\i~''
3/110 120^3^
Xi 4
But from (1) we have y^ — 10 = 30 Xi and hence
^Zi? = 30.
Xi
This proves that every point whose coordinates {x^, y^ satisfy
the relation 2/ = 30 a; + 10 is on the straight line AB.
Conversely, every point on the straight line AB has coordinates
{xij yi) which satisfy the relation y = SO x \ 10.
For, from the figure, we have
yi10 ^120
Xi 4
whence
yi=S0x^\10.
* Extended beyond B, if Xi > 4.
** Observe that QP and CB are measured in different units from AQ and
AC. But the ratio of two linesegments is independent of the unit in which
they are measured.
Ill, § 49] LINEAR FUNCTIONS 67
The straight line AB (extended indefinitely beyond B) then
gives a complete representation of the function 30 a; + 10, at
least for positive values of x (negative values of x have no
meaning in this problem). Every pair of corresponding values
of X and y gives rise to a point on AB, and eveiy point of AB
has coordinates which are corresponding values of x and y. By
virtue of this fact the line xiB is called the graph of the function
30 a; 4 10, or the locus of the equation y = 30 x { 10 referred to
rectangular coordinates ; Avhereas the equation ?/ = 30 a; + 10 is
called the equation of the line AB.
Uses of the Graph. The graph just discussed exhibits
vividly to the eye several properties of the function 30 a; 4 10.
(1) The function steadily increases as x increases. This
corresponds to the fact that the longer the train moves east
ward, the greater is its distance from Buffalo.
(2) Corresponding to every positive value of a;, there is a
unique value of y. From the graph find y when x is 4.
(3) Corresponding to every positive value of y (greater than
10) there is a unique value of x. What is the value of x when
.vis 160?
(4) The last consideration means that x is also a function of
y. Explicitly we have
y = 30 a; + 10,
whence
2/  10 = 30 a;
and
x=.y^^.
30
•It is left as an exercise to draw the graph of the function
y10
^='"^0~
by assigning values to y and computing the corresponding
values of x. Compare the result with the graph in Fig. 30.
68 MATHEMATICAL ANALYSIS [III, § 49
Rate of Change of a Functiox. Before leaving this special
case to consider a more general problem, we shall use it to illus
trate a very important conception connected with a function.
We have noted that when a; = 0, y = 10. Starting from this
initial value, as x increases from the value 0, the value of the
function, i.e. _?/, changes (in this case increases). It is often of
the greatest importance to know how the increase in the func
tion y is related to the increase in x. As x increases from
to 1, y increases from 10 to 40 ; i.e. a change in x of one unit
produces a change in ?/ of 40 — 10 or 30 units. The relative
change is then ^^2^, or 30. As x increases from x = to x = 2, y
changes from y z= 10 to y= 70, or by 60 units, and the relative
change is again 30.
Let us see what the situation is in general. Let Xi be any
particular value of x and yi the corresponding value of y ; then
suppose that Xo is any other (subsequent) value of x and ?/2 the
corresponding value of ?/. The change in x is evidently X2 — Xi
and the corresponding change in y is y2 — 2/1 We seek the
value of the ratio
X^Xi
We have from the data of the problem
2/2 = 30 X2 + 10
and
y^ = 30 a^i h 10.
Subtracting we get
2/22/1 = 30(^2 aJi)
and hence
^23/1^30,
0^2 — a?!
We see then that the ratio of a change in the function 30 a; +10
to the corresponding change in x is constant and is equal to
Ill, § 50]
LINEAR FUNCTIONS
69
the speed of the train. We shall see presently that in any
function of the first degree in x, the ratio of a change in the
function to the corresponding change
in X is constant.
Geometrically this result expresses
the familiar proportionality of homol
ogous sides of similar triangles. By
reference to Fig. 31 we may readily
verify that the terms 2/2 — 2/i ^^^^ ^2 — ^1
represent the vertical and the hori
zontal sides of a right triangle whose
hypotenuse is on the line AB. The fact that the ratio
(2/2 — 2/i)/(^2 — ^1) is constant, i.e. always equal to 30, simply
corresponds to the obvious fact that any two such triangles,
no matter at what place they are drawn, or how long their
sides are taken, are similar.
\ l^'
IV
inn
~ A t"
^^ V
^ i
yfr T^
eo ~" ,> 1:1 1 1 \\r\ 1
bU ^^
I ISI
40 P .^
4^ :
20i'^V Xn
= •^1 + 
+
^ 2
3 i
Fig. 31
50. Change Ratio. The ratio
X2 — X1
is called the change ratio (or sometimes the difference ratio) of
the function. The difference X2 — x^ is often denoted by Ace, and
the corresponding difference y^ — 2/1 by Ai/.* The change
ratio may then be written A?//Aiw. Explicitly, by definition,
we have the following equalities :
change ratio = ^ = fclli^ = change in y .
Ajt X2 —Xi corresponding change m x
The preceding considerations suggest the theorem :
* A is a Greek capital letter corresponding to our D and called delta ; it is
used because d is the initial of the word ** difference." " Ax," is then merely
an abbreviation for "difference of the x's " or '* change in z " and " Ay " for
" difference of the y's " or " change in y."
70 MATHEMATICAL ANALYSIS [III, § 50
If the change ratio of a function is constant, the graph of the
function is a straight line; and conversely.
The truth of this theorem is already sufficiently indicated in
case 2/2 — Vi ^^d ^2 — ^1 3.re both positive numbers. In formu
lating a general proof we must keep in mind that y^ — 2/1 ^^d
X2 — Xi may be either positive or negative and that these dif
ferences represent directed segments. The proof of the theorem
in general will appear presently.
EXERCISES
1. Discuss fully the graph of the function y = 2 a; + 3. Prove that the
graph is a straight line. Express x as a function of y. Find the change
ratio and show that it is constant.
2. Proceed as in Ex. 1 for each of the functions :
(a) 5x42, {h)x+\2, (c) 3.2a: + 8.4.
3. Prove that the change ratio for the function y = mx + 6 is m.
4. A steamer 150 miles east of Toledo starts to travel west at a uniform
rate of 15 miles per hour. Express its distance y east of Toledo at the end
of X hours. Draw the graph of the function and prove that it is a straight
line. Does the distance y increase as x increases ? Calculate the change
ratio and show that It is constant. What is the significance of the
negative sign ? At what time is the steamer 10 miles east of Toledo ?
When does it reach Toledo ? How are the last two results shown in the
graph ? What is the significance of the graph that extends below the xaxis ?
5. Give examples, drawn from your experience, of functions which
(a) increase as the variable increases ;
(6) decrease as the variable decreases.
6. Consider the function y ■= x^. Calculate the corresponding values
of y when a; = 0, 1, 2, 3, 4, 5. Plot the corresponding points and observe
that they are not on a straight line. Calculate the change ratio of this
function for x = and Ax = 1, 2, 3, and observe that it is not constant.
7. The cost of printing certain circulars is computed according to the
following rule. The cost for the first one hundred circulars is $ 2 and
for each succeeding one hundred $0.50. Express the cost y in dollars of
X hundred circulars. Draw the graph of the function and determine
from the graph the cost of printing 475 circulars. What does the change
ratio of the function y express in this case ? Ans. y = ^ x + f .
Ill, § 51] LINEAR FUNCTIONS 71
51. The General Linear Function mx h b. The change ratio
of every function of the form mx + b is constant.
Proof. Let (x^ , yi) and (ajg , 2/2) be any two pairs of cor
responding values. Then
2/1 = w^i + & and 2/2 = mx2 \ h ;
hence , ^
2/2  .Vi = m (icg  xi),
I.e.
Conversely, if the change ratio of the function y of x is con
stant and equal to m, the function has the form y = mx + b.
Let (xi, 2/1) be a particular pair of corresponding values and
(x, y) any other pair of corresponding values. By hypothesis
the change ratio is equal to m ; i.e.
X — Xi
or
y = mx — mxi f 2/1 ;
but — mxi 4 2/1 is a constant, say b. Hence
y = mx h b.
Hookers law affords an excellent illustration of the above theorem.
This law states that the length 1/ of a piece of wire under tension is equal
to its original length 6, plus the stretch, which is proportional to the force
X causing it. Thus, y z=l h + mx.
This law may also be stated simply by saying that the change ratio of
the length y^ with respect to the pull a:, is constant.
The preceding considerations lead to the following theorem.
Theorem. If a function y of a variable x is such that any
change in the value of the function is always equal to m times the
corresponding change in the variable, the function y is given by
a relatian of the form y — mx f b, and, conversely, in any func
tion of this form any change in y is always m times the corre
sponding change in x.
72
MATHEMATICAL ANALYSIS
[III, § 52
52. The Graph of a Linear Function. Let Pi{xi , y^) be any
point on the graph of a linear function (Fig. 32). From Pj draw
to the right a positive horizontal segment P1Q2 equal in length
to X2 — Xi , i.e. Ax. Through Q2 draw a vertical segment and let
it meet the graph in the point Po • The segment Q2P2 is equal
to 2/2 — yi? i'^ ^y, and is positive if P2 is above Pi (Fig. 32 a)
}' . __
Pz
Pi
Q% Qs
la)
X
Fig. 32
Pi
Q2 Qb
A
A
(6)
and negative if P2 is below Pi (Fig. 32 6). Now let us take
another positive change Ax = x^ — Xi, represented by P] Q.^ and
the corresponding change Ay — y^ — 2/1 represented by Q3P3 .
If the change ratio is constant, then (1) either Pg and P3 are
both above Pi or they are both below Pj , according as the given
constant is positive or negative; and (2) the triangles P1Q2P2
and P1Q3P3 are similar. Therefore the points P1P2P3 are on a
straight line, if and only if the change ratio is constant.
Theorem. The graph of any function of the form y = mx f b
is a straight line.
To draw the graph of such a function we need, therefore,
merely to plot two points of the graph and draw the straight
line through them.*
* While two points are sufficient to determine the line completely, it is
desirable to find a third point as a cheek on the other two. Moreover, it is
advisable to take the points as far apart as convenient. Why ?
Ill, § 54]
LINEAR FUNCTIONS
73
Y
C
n
B
y^
D
,^
1
y
IM
2 X
Fig. 33
In the figure this ratio
Example. Draw the graph y = ^x \ 2. We notice that (0, 2) and
(4, 14) are two points on the graph. The line joining these two points is
the required line. Check by plotting a third point.
53. The Slope of a Straight Line. The graph of the func
tion y = mx + b may be obtained by observing that x = Oj
y ^b and x = 1, y = m \ b are
two pairs of corresponding values
of X and y. In the adjoining
figure (Fig. 33) we have plotted
the two points B (.0, b) and
0(1, m + b) on the assumption
that both of the quantities b and
m are positive numbers. The
change ratio, as we have seen, is m.
is DC/BD.
Now suppose that b remains constant and that m takes on
successively different values. Under the hypothesis that b
and BD remain fixed, the points B and D would remain fixed
and the point C would move up or down on the vertical line
through D, according as m increases or decreases. The line
BG would then rotate about the point B, becoming steeper if
m is increased and less steep if m is decreased. The change
ratio m then measures the steepness of the line. The term
change ratio applies to the function mx \ b ; when applied to
the straight line y = nix f 6, it is called the slope of the
straight line.
54. Remarks Concerning the Slope of a Line. We as
sumed in the last section that both b and m were positive
numbers. Let us now suppose that b is still positive, but that
m is negative. Observe that in the preceding figure MC
= MD + DC = b\m. Eecalling that the relation MC= MD
+ DC holds universally for any three points M, D, C, on a
74 MATHEMATICAL ANALYSIS [III, § 54
line (Art. 35), the interpretation of a negative m, i.e. DC, is
that the point C is below, the point D. (Cf. also § 52.) A
negative value of m then merely causes the line to slope
downward in going from left to right, while, as we have seen, a
line with a positive m slopes upward. When m = 0, the line
is parallel to the a^axis. Indeed the equation y = mx + h be
comes, for the value m = 0, the equation y = h. This equation,
when interpreted as a function of x, means that for every value
of X, the value of ?/ is h \ the graph of such a function is ob
viously a straight line parallel to the avaxis. Since a change
in X in this case produces no change in y, the change ratio is
zero. Finally, if h is negative, nothing is changed except that
the point B is below the origin 0. A positive m still indicates
an upward slope and a negative m a downward slope, in pass
ing from left to right.
The number h, we have seen, represents the segment from
the origin to the point in which the line cuts the yaxis. This
segment is called the yintercept of the line. Similarly, the
segment from the origin to the point in which the line cuts the
icaxis is called the xintercept of the line.
We have then the following results :
Tlie straight line represented by the equation y = mx + b has a
slope equal to m a7id a yintercept equal to b. In passing from left
to right, the straight line slopes downward ifm is negative and up
ward if m is positive; if m is zero, the line is parallel to the xaxis.
In the terminology of functions we have :
The linear function mx f b is an increasing function of x {i.e.
the function increases as x increases) if the charige ratio m is pos
itive, and a decreasing function of x (i.e. the function decreases
as x increases) if m is negative. It is a constant function if m
is zero.
Ill, § 55]
LINEAR FUNCTIONS
75
55. Examples of Linear Functions. Example l. On a Fah
renheit thermometer the freezing point of water is placed at 32°, the boil
ing point at 212°. On a Centigrade thermometer the freezing point is
at 0°, the boiling point at 100^. Express the temperature of y° Fahren
heit as a function of y° Centigrade.
Solution : y = 32 when a; = 0. Also the range of temperature from
the freezing point to the boiling point of water is 212°32° or 180° F. while
it is 100° C. Therefore it follows that an increase of 1° C. is equivalent
to an increase of  of a degree F. Now as the temperature increases from
0° to ic° C. the change in the number of degrees is x. This change in
temperature is equivalent to an increase from 32° to y° F. The change
in the number of degrees is then
?/32i=x, or y =x + 32.
As a check we may observe that, when x = 100, the formula gives y = 212,
as it should. Are negative values of x admissible ? Figure 34 represents
the graph of this function. It was drawn by using the points A (30, 22),
B (100, 212). [Why is it desirable to choose points so far apart?]
This "graph may be used to read off without computation the approximate
temperature in F. for a given temperature in C. For example, to x = 22
corresponds y = 72, approximately. Therefore 22° C. is equivalent to
about 72° F. By computation we find that y = ll.Q.
■
1
'
'
"
'
. .?
zko
i
^
' /
^
t
/
'4
1
/
js
/
^
/
*
;5
1
Si)
2^
/
y
/\
/
/
/
/
/
/
/
.i
5
p'fl
no
7 a
.
/
I)
e.
rj.
es
4
iti
I'
ii.(.e\
r
/
A
'A
._
.
1 1
Fig. 34
76 MATHEMATICAL ANALYSIS [III, § 55
Example 2. A bar of iron 3 ft. long at 60"^ F. will expand or con
tract if tlie temperature increases or decreases. The increase in length is
proportional to the increase in temperature (physical law). More pre
cisely, an increase of 1° F. produces an increase of 0.0000027 ft. In this
case we have m = 0.0000027. If y represents the length at x° F., we have
?/ = 3 + m(x60).
Does this relation hold also when x < 60 ? Why ? Can you draw the
graph ?
We shall now give an example in which m is negative.
Example 3, An aeroplane starts 200 miles eas^of Chicago and travels
towards Chicago. Express its distance y from Chicago in miles at the
end of t hours, if the aeroplane moves at the rate of 82 miles per hour.
Solution : According to the data the distance from Chicago is de
creasing at the rate of 82 miles per hour, i.e. m =— 82. Therefore,
y2Q0=82t, or y = S2 t \ 200.
Draw the graph. What is the significance of a negative y (e.g. when
^ = 5)? When does the aeroplane reach Chicago ? When is it at a point
63 miles east of Chicago ? When is it 52 miles west of Chicago ? How
could these questions be answered from the graph alone ?
EXERCISES
1. On a Reaumur thermometer the freezing point of water is at 0°,
the boiling point at 80°. Express the temperature in degrees Fahrenheit
in terms of the temperature in degrees Reaumur. Draw the graph and
show how it may be used.
2. Is there any temperature whose measures in the Fahrenheit and in
the Centigrade scales are equal ? Answer by computation. How could
the result be found graphically ?
3. A cistern that already contains 300 gallons of water is filled at the
rate of 50 gallons per hour. Show that the amount of water y in this cis
tern at the end of x hours is y = 50 x } 300. Draw the graph and discuss.
How would the function be changed, if the cistern were being emptied
at the rate of 50 gallons per hour ?
4. A tank contains 16 gallons of water. A faucet is opened which
admits 4 gallons per minute. Express the amount, w, of water in the
tank at the end of t minutes. Draw the graph. Do negative values of t
have any significance ? When will the tank contain 37 gallons ?
Ill, §56] LINEAR FUNCTIONS 77
5. A tank containing 37 gallons of gasolene is emptied at the rate of
5 gallons per minute. Express the amount of gasolene in the tank at the
end of t minutes. Draw the graph. When will the tank be emptied ?
For what range of values of t has the function any significance ?
6. On a certain date a man has 1 5 in the bank. At the end of every
week he deposits $3. How much money has he in the bank at the end
of X weeks ? Draw the graph of this function. How is the rate of in
crease shown in the graph ?
7. On a certain day a man has 1 100 in the bank. At the end of
every week he draws out '$5. How much money has he in the bank at
the end of x weeks ? Draw the graph of this function. How is the rate
of decrease shown in the graph ?
8. In experiments with a pulley, the pull P in jjounds required to lift
a load L in pounds, was found to be P = 0.15 L + 2. Plot this relation.
How much is P when L is zero. How much is P when L is 10 lbs.?
9. If h represents the height in meters above sea level, and h the
reading of a barometer in millimeters, it is known that b — k ■{■ hm, where
k and m are constants. At a height of 110 meters above sea level the
barometer reads 750 ; at a height of 770 meters it reads 695. What
equation gives the relation between b and Ji ? Draw the graph of this
equation and from the graph determine h when b = 680.
56. Linear Interpolation. The fact that the change Ay in a
linear function y is proportional to the change Aa; in the variable
X makes it possible to interpolate readily. For example, if we
know that ?/ is a linear function of x, and that y = 432.50 when
X = 32.0 and that y — 436.90 when x = 33.0, we can calculate
mentally the value of y when x = 32.3. For we know that in
this case Ay = 4.40 when Ax = 1.0 ; hence A?/ = 4.4 x 0.3 = 1.32
when Ail' = 0.3. Hence y is 433.82 when x = 32.3. This pro
cess is known as linear interpolation. Why would this process
not apply directly to functions that are not linear ?
EXERCISES
Assuming that y is a linear function in each of the following cases
compute the values of y indicated.
1. When a; = 10, ?/ = 50; when x = 14, y = 90; when a; = 11, y = ?
2. When x = 2.4, ?/ = 9.8 ; when x = 2.5, y = 8.Q; when x = 2.42, y=?
78
MATHEMATICAL ANALYSIS
[III, § 57
57. Graphic Solution of Problems. Whenever we know
at the outset that the solution of a problem is going to depend
on the consideration of one or more linear functions, we can
often solve the problem graphically without determining these
linear functions analytically. Such a method is advantageous
whenever the computation is difficult or tedious and when
great accuracy is unnecessary. In order to decide whether
the functions involved are linear or not, we usually have re
course to the theorem (§ 51) that, whenever the change in
the function is proportional to the change in the variable, the
function is linear. This is true, for example, in all cases of
motion at a constant speed on either a straight or curved
path ; the distance is then a linear function of the time.
The following example will serve to illustrate the graphic
method of solution.
Example. At 7 a.m. a man starts to go up the 7mile carriage road
of Mt. Washington. At 9 o'clock he passes a party of ladies coming
down. He reaches the top at 10 o'clock and, finding no view, he immedi
ately sets out on the return trip, which takes 1 f hrs. As he reaches the
hotel from which he started he notices the party of ladies just arriving.
At about what time did the ladies leave the top, assuming that the man
kept up an approximately constant rate of speed on the way up and the
ladles on the way down ?
To solve the problem, we represent on a horizontal axis the time, mark
ing the hours 7, 8, 9, 10, 11, 12 and on the vertical axis the distances
from the hotel at the foot of the
t> m — I I ' I — I.I I I I I
mountam. The graph' of the man
going up the mountain is a straight
line starting at O (at 7 a.m. he was
at the hotel) and ending at a point
A representing 10 o'clock and 7
miles from the hotel. Regarding
the ladies, we know that the graph
of their descent is also a straight
line. At 9 o'clock they were the
same distance from the hotel as
the man. The point B on the line OA, corresponding to 9 o'clock,
/>
A
N
/
\
>v
/
>
iP
/
/
\
/
\
z
/
\
\
b£
07
9 10
Fig. 35
11 12 houna
Ill, § 58]
LINEAR FUNCTIONS
79
is then one point of the ladies' graph. Another point is the point C
at distance from the liotel at n:45. The line BC is then drawn
and extended to Z>, representing 7 miles distance from the hotel. It is
seen that the ladies started at about 7:30. How far was the man
from the top when he met the ladies ?
58. Sum of Two or More Functions. Let mix h h^ ,
m^x\h2 , '", m^x f 6^ be any k linear functions of x. The sum
of these functions is (miX + 6]) \{m2X [ 62)+ ••• H(^fc» + &*)
and this is equal to
(mi+W2f ••• +mj)x+{hi 462H h &jk)>
which is again of the form mx + h. The result may be stated
as follows : The sum of any number of linear functions of x is
itself a linear function of x.
Example. An empty tank is being filled by a faucet supplying 2
gallons of water per minute. After this faucet has been running 5
minutes a second faucet is turned on which supplies water at the rate of
3 gallons per minute. When the two faucets have been running to
gether for 6 minutes, an outlet is opened, but both faucets continue to
Fig. 36
run. If the tank is empty at the end of 32 minutes, counted from the
start, draw a graph representing the amount of water in the tank at any
instant. Find approximately the rate of flow from the outlet, which may
here be considered constant.
80 MATHEMATICAL ANALYSIS [III, § 58
We shall represent minutes on the horizontal scale and gallons of water
in the tank on the vertical scale. The increase of water due to each
faucet is at a constant rate, and the decrease when the outlet is opened is
also at a constant rate. The amount of water in the tank due to each
cause separately is, therefore, a linear function of the time, and their
algebraic sum is also a linear function of the time. The first faucet begins
at < = (when w, the amount of water in tank, is 0) to supply water at
a uniform rate which would supply 40 gallons in 20 minutes. The
amount of water in the tank due to the first faucet almie would then be
represented at any instant by the straight line OA joining the points
O (0, 0) to A (20, 40). The second faucet begins when t = 5 to
supply water at the rate of 30 gallons in 10 minutes. If this second
faucet were operating alone, the water supplied by it at a given instant
would be represented by the straight line joining B (5, 0) to C (15,
30). In the actual problem from the instant i? = 5, the two faucets are
running simultaneously. The sum of the two functions is then rep
resented by the linesegment DE, where Z> = (5, 10) and £' = (10, 20 + 15)
= ( 10, 35). This line may be obtained graphically from the figure. When
t = 11, a new factor enters, which reduces the amount of water in the
tank to zero at i = 32. You may now finish the discussion. The required
graph is the broken line ODHI. What would be the effect on the graph
if one or both faucets were turned off at < = 20, the outlet remaining open ?
EXERCISES
1. A man on horseback rides from a place J. to a place 5, 15 miles
distant, in 2 hours. When he is 4 miles from A, he passes a lady walk
ing in the same direction. The man remains at B \ hour and then
returns to A on foot. After walking 1 hour, he meets the lady on her
way to B. If the man walks at the rate of 3 miles per hour, find the
rate at which the lady is walking and at what time she left A.
2. A man starts at A to walk through B to a place C. At the same
time a second man starts to walk from B to C. The first man reaches
B in 1^ hours, while the second man has only walked f as far in this
time. In how many hours will the first man overtake the second ?
3. Represent graphically on the same drawing the motion of the hour
and the minute hand of a clock and use the drawing to determine ap
proximately at what time the two hands are in the same position.
[Hint: The hands are together at twelve. Lay off the hours from 12
(or, 0) to 12 on the horizontal axis and the angles in degrees that either
Ill, § 59]
LINEAR FUNCTIONS
81
hand makes with the 12 o'clock position on the vertical axis. Each liand
moves at a constant angular speed. The graph of the hour hand is then
a straight line joining the points
(0, 0), (12, 360). The minute hand
goes from to 360 in 1 hour. The
graph during the first hour is then a
straight line joining (0, 0) to (1, 360).
At 1 o'clock the graph begins at
(1, 0) and goes to (2, 360) and so on.]
4. At what time between five and
six o'clock are the hands of a watch
together ?
5. At what time between two and three o'clock are the hands of a
watch opposite to each other ? At right angles ?
6. At what time between four and five o'clock are the hands of a
clock at right angles ? (Two solutions.)
7. A and B start to walk towards each other from two towns 15
miles apart. A walks at the rate of 3 miles per hour but rests one hour
at the end of the first 6 miles. B walks 4 miles per hour but rests two
hours at the end of the first 4 miles. In how many hours do the two men
meet ?
8. Two men can do a certain piece of work in 12 and 15 days re
spectively. After the first man has worked 3 days alone, the two men
finish the work. How long do they work together ? Ans. 5 days.
9. A messenger boy riding a bicycle at the rate of 9 miles per hour
is sent to overtake a man on horseback riding 6 miles per hour. How
long will it take the boy to overtake the man if the man had a start of
4 miles ?
59. Explicit and Implicit Functions. We have hitherto
considered functions which were defined explicitly by an
expression involving the variable. Thus the relation between
y° Fahrenheit and x° Centigrade was expressed by the relation
Now let us consider the equation 2x — Sy{7 = 0. This
equation also defines a functional relation between two vari
82 MATHEMATICAL ANALYSIS [III, § 60
ables. To every value of x corresponds a definite value of 2/,
and, conversely, to every value of y corresponds a definite
value of X. But, the equation does not express one of the
variables explicitly as a function of the other. In fact the
form of the equation gives no indication which of the variables
is to be considered as the independent variable and which as
the function. Such a relation is said to define a function
implicitly.
From such an implicit relation we can derive the expression
of either variable as an explicit function of the other. Thus
from 2iB— 32/f7 = follows at once
2/ = l^+i and x=:f2/f
The first of these equations expresses y as an explicit function
of X, and the second expresses x as an explicit function of y.
60. The General Equation i4x h 5y + C = 0. Any linear
relation between two variables x and y can be written in the
form
(1) Ax^By\C^Qf.
For example, the relation just discussed in the preceding arti
cle is obtained from this general relation by placing ^ = 2,
5 = — 3, (7=7. Equation (1) always defines ?/ as a linear
function of x^ except when 5 = 0. In this case the term in
volving y drops out and the equation reduces to Ax + C = 0,
and we cannot speak of 2/ as a function of x.
But, if 5 :^ 0, we have By = —Ax—C^ or
A C
y = — ^ — J
^ B B'
which is of the form y = mx f h. Hence we conclude :
Any equation of the form Ax + By \ C =0 defines y as a
lineal' function of x for all values of A, B, C except B = 0.
Ill, § 61] LINEAR FUNCTIONS 83
61. The General Equation of the Straight Line. It follows
from the result of the last section, that the locus of the equa
tion
when interpreted geometrically in rectangular coordinates, is
a straight line, except perhaps when B = 0, when the equa
tion takes the form Ax \ C = 0. In this case, if A = also, the
equation reduces to (7 = 0, and it completely disappears. If A
is not zero, we may solve the equation for x and obtain,
G
. = ,
or
« = a constant.
Now, the locus of a point whose abscissa is constant is a line
parallel to the ?/axis and at a distance equal to the constant
from it. Thus the locus of a; = — 3 is a line parallel to the
ySixis, and three units to the left of it.
The case JB = is not then an exception, and we have the
following theorem.
Every equation of the form Ax ■\ By \ C = 0, when repre
sented geometrically by means of rectangular coordinates, repre
sents a straight line. If B = 0, the line is parallel to the yaxis ;
1/^ = 0, the line is parallel to the xaxis; if (7=0, the line passes
through the origin.
Prove the last two statements of this theorem.
We may also state the following theorems.
Every straight line in the plane may he represented by an equa
tion of the form Ax j By \ C = 0.
The loci of AxiBy{ C=0 and k (Ax \ By \ C) =
(k =^ 0) a?'e identical.
The proofs of these theorems are left as exercises.
84 MATHEMATICAL ANALYSIS [III, § 62
62. Analytic Geometry. We have thus far used the notion
of coordinates to give a geometric interpretation to algebraic
relations. It is possible to reverse the process and use the
connection established between algebra and geometr}^, for the
study of geometry. This method of studying geometry by
algebraic means is called analytic geometry. In the following
sections we proceed to develop certain analytic methods
applicable to the straight line. The results are, in a large
measure, merely a restatement from a different point of view
of the results already obtained.
63. Straight Lines. We have already seen that the graphs
of equations Ax \ By \ C = and y = mx + & (§ 52), when
represented by means of rectangular coordinates, are straight
lines. In § 60 we saw that the first of these equations could
be put in the form of the second, provided 'B ^ 0. Thus when
an equation of the form Ax { By \ C = is solved for y, the
coefficient of x is the slope, and the constant term is the yintercept.
The slope of the line connecting the two points Pi(iCi, 2/i),
A(aJ2,2/2) is (§§5153)
We see geometrically that a line is determined when we
know its slope and a point on the line. To determine the
equation of this line, if {x^, y^) is the given point and m the
given slope, we proceed as follows. Let {x, y) be any variable
point on the line. Then, equating slopes, we have
X — Xi
that is,
yy,= m{x X,)
is the required equation.
Ill, § 64]
LINEAR FUNCTIONS
85
It is left as an exercise to prove that the equation of the
straight line through the two given points (x^, y^), (ajg, 2/2) is
if Xi =^ X2.
2/yi = ^'_^' (^^i)>
64. Parallel Lines. In Fig. 37 let (1) and (2) be two
parallel lines with slopes mi and mg. Construct the positive
segments PiQi and P2Q2 from the points P^ and P2 on lines (1)
Fig. 37
and (2) respectively, and complete the right triangles P^QiRi
and P2Q2R2' We then have
m
, = ^1^ and m2 = ^'^'
If the lines are parallel, QiRi and Q2R2 are either both positive
or both negative ; m^ and mg have then the same sign. They
have the same magnitude since the triangles P^QiRi and
P2Q2R2 are similar. Hence,
If two lines are parallel, their slopes are equal, i.e. m^ = mj.
Conversely, if the slopes of two lines are equal, the lines are
parallel.
The proof of this statement is left as an exercise.
86
MATHEMATICAL ANALYSIS
[III, § 65
7
V
{
9
A
y
\
\
X
Fig. 38
vertical line B1R2 .
65. Perpendicular Lines. In
Fig. 38 let (1) and (2) be two
perpendicular lines with slopes
mi and rriz and let the units on the
two axes be equal. From the
intersection P of the two lines
construct the positive horizontal
segment PQ of any convenient
length. Through Q draw the
We then have
m, =
^Ma„dm, = «^
PQ
PQ
Therefore the slopes have opposite signs. Why ? Also from
the right triangle R.R^P we have P^ =  QE^ \ •  QR2 1. There
fore * mim2 = — 1 and
m, =
That is, if the units on the coordinate axes are equal , perpendicu
lar lines have slopes ivhich are negative reciprocals of each other.
Conversely, if the slopes of two lines are negative reciprocals of
each othery the lines are perpendicular, provided the units on the
coordinate axes are equal. The proof of this statement is left
as an exercise. Why is it necessary to assume the units equal ?
66. Illustrative Examples. Example 1. Find the equa
tion of the straight line through the point (4, 7) and having
the slope — 2.
* This proof presupposes that neither mj nor mg is zero, i.e. the lines are
not parallel to the coordinate axes, and the result obtained does not apply to
such lines. However, two lines parallel to the x and y axes have equa
tions of the form y = a constant and a; = a constant, respectively, and henco
can be recognized at once.
Ill, § 66] LINEAR FUNCTIONS 87
We have at once from § 63, ?/ — 7= — 2 (ic — 4)
or
2 a; f 2/  15 = 0.
Example 2. Find the equation of the straight line through
the points P^{2, 4), P^{ 5, 6).
rru 1 64 2
The slope m =  —  —  =
— 5—2 7
From § 63 the equation of the line is y ~ 4:= — ^ (x — 2) ov
7y + 2a;32 = 0.
Example 3. Express the temperature measured by y° Fah
renheit as a function of fl;° Centigrade.
We know that when y = 32, x = 0: i.e. Pi(0, 32) is a point
on the graph. In the same way we have PaC^^O, 212) a point
of the graph. Therefore the equation of the line connecting
these points is
' 21232 ^ y32
100  a; 
or
2/ = 9 a; + 32 (See § 55, Example 1.)
Example 4. Find the equation of the straight line
through the point (2,5) and parallel to the line 2 y + 4 ic — 5
= 0.
The slope of the given line is — 2 (§ 63). Therefore
the equation of the required line is y{5= — 2(x — 2) or
2a; + 2/ + 1 = 0.
Example 5. Find the equation of the straight line through
the point (1, — 2) and perpendicular to the line 3x — y \2 = 0.
The slope of the given line is 3. Therefore the slope of
the required line is — ^ (§ 65). The equation of the required
line is2/f2 = — i(a; — l)ora;f32/45 = 0.
88 MATHEMATICAL ANALYSIS [III, § 66
EXERCISES
1. What is the meaning of the constants m and h in the equation
y = mx + b?
2. What is the effect on the line y = mx + & if 6 is changed while m
remains fixed ? If m changes when b remains fixed ?
3. Describe the effect on the line y — yi = m(x — xi) if m changes
while xu y\ remain fixed : also describe the effect if Xi, yi, vary while m
remains fixed.
4. What is the equation of the line
(a) whose slope is 3 and whose ^intercept is 2 ; Ans. y = 3 x + 2.
(6) whose slope is 4 and whose yintercept is — 3 ;
(c) whose slope is and whose ^/intercept is — 1 ;
(d) whose slope is and whose yintercept is ?
5. Describe the positions of lines (c) and (d) in Ex. 4.
6. Define " yintercept of a line." What is meant by the "xinter
cept"?
7. For each of the following lines give xintercept, yintercept, and
slope :
(a) 2x3?/ = 7. ^ns. I; I; f. (c)2xy + 5 = 0.
(6) X + y  2 =: 0. ((?) 4 X + 2/ = 0.
8. Is a straight line determined if we know its intercepts ? Try
each of the equations 2 x — y = 4 and 2 x — y = 0.
9. Find the equation of the line joining the two points (2, 1) and
(—3, 1); of the line joining the points (4, 2) and (4, —3).
10. Which of the following lines are parallel ?
(a) 2 X  2/  4 = 0. (c) 4 X  2 ?/  1 = 0.
(6) y + 2 X + 3 = 0. (d) 2 y + 4 X + 5 = 0.
11. Are the points (1, 5), ( 1, 1), (2, 6) on the line y = 2 x + 3 ?
12. What is the equation of the line which is parallel to y = 2 x + 3
and passes through the origin ? perpendicular to y = 2 x + 3 and passes
through the origin ?
13. Determine k so that
(a) the line 2x + 32/ + A; = shall pass through the point (0, 1) ;
Ans.  3.
(b) the line 2xf3y4* = shall have a yintercept equal to 2 ;
{c) the line 2x + 3y + ^ = shall have an xintercept equal to 5.
Ill, § 66] LINEAR FUNCTIONS 89
14. Find the equations of the sides of the triangle whose vertices are
(3,4), (1, 2), (4, 5).
Ans. x2y + 5 = 0; 9xly + l=0 7x3i/+13 = 0.
15. Find the equations of the sides of the quadrilateral whose vertices
are (2,1), (3, 1), ( 2, 4), (1, 7).
16. What intercepts does the line through the points (2, — 7) and
(4, — 5) make on the axes ?
17. Find the equation of the line which passes through the point
(4, — 2) and whose slope is 6.
18. A line has the slope 2 and passes through the point (—1, 2).
What are its intercepts ?
19. What is the equation of the line which passes through (—6, 5) if
its yintercept is — 3 ? Ans. 8a: + 5?/ + 16 = 0.
20. Write the equations of the lines which make the following inter
cepts on the X and !/axes.
(a) 2 and  4 ; (b)  7 and  3 ; (c) 4 and 5 ; (d) and 0.
21. If the X and yintercepts of a line are a and b, prove that the equa
tion of the line can be written iii the form
a b
[This equation is called the intercept form of the equation of a straight
line.]
22. Solve Ex. 20 by using the result of Ex. 21. Does the formula hold
in Ex. 20, (d) ? Explain.
23. Find the equation of the straight line through the point (4, — 6)
parallel to the line 2x— y{7 = 0; through the same point, perpendicu
lar to the line 2x — y + i = 0. Ans. y = 2x— IS; 2y = — x — Q.
24. Prove that the lines Ax + By \ C =0 and Ax { By + D = are
parallel. State this theorem in words.
25. Prove that the lines Ax \ By + (7 = and Bx—Ay^D = are
perpendicular. State this theorem in words.
26. Prove that the lines Ax \ By + = and Mx h Ny \ P = Sive
perpendicular if and only if AM {■ BN = 0.
27. Show that the points ( 8, 0), (4,  4), (4, 4), and (4,  4)
are the vertices of a trapezoid.
28. The Reaumur thermometer is graduated so that water freezes at 0°
and boils at 80°. Find the equation of the line that represents the read
90
MATHEMATICAL ANALYSIS
[III, § 66
ing B of the Reaumur thermometer as a function of the corresponding
reading C of the Centigrade thennometer.
29. A printer asks 75 cents to set the type for a notice and 3 cents per
copy for printing. The total cost is what function of the number of
copies printed ? Draw the graph of this function.
30. Express the value of a $ 1000 note at 6 % simple interest as a
function of the time in years. Is this a linear function ?
31. A cistern is supplied by a pipe that supplies water at the rate of 30
gallons per hour. Assuming that the amount A of water in the cistern
is connected with the time ^ by a linear relation, find this relation if
^ = 1000 when t = 10. What is A when t = 0?
32. In stretching a wire it is assumed that the elongation e is con
nected with the tension t by means of a linear relation* Find this rela
tion if i = 20 lb. when e = 0.1 in. and t = 60 lb. when e = 0.3 in.
67. Systems of Straight Lines. An equation of the first
degree in x and y, and containing an arbitrary constant, repre
sents in general an infinite number of
straight lines. For the equation will
represent a straight line for each value
of the constant. All the lines repre
sented by an equation of the first
degree containing an arbitrary con
stant are said to form a system of
lines. The arbitrary constant is called
the parameter of
the system.
Thus the equa
tion yz= — 3x{b represents the system
of straight lines with slope — 3. (See
Fig. 39.) The equation y — 2 = 7n{x — l)
represents the system of straight lines
through the point (1, 2).* (See Fig. 40.)
* It represents every line of this system except the one parallel to the
yaxis. Why ?
Fig. 39
1
c
: i^.Zt
^v
i X
  ^:,:
[V i y
'
sA' *"
J^S
rt
Q"~ ^ V
:.^^^\n.
i :: 5^
ii 1i^ X
::: dZ
±i: :±
Fig. 40
lU, § 68] LINEAR FUNCTIONS 91
68. Pencil of Lines. All the lines in a plane which pass
through a given point are said to form a pencil of lines. The
point is called the center of the pencil. 11 Ax + By ~\ C = 0,
and A'x + B'y \ C' = are any two lines of the pencil, then
(3) (Ax f % 4 0)h k(A'x + B'y + 0')= 0,
where k is an arbitrary constant, represents a line of the
pencil. This is true because the equation (3)
(a) is of the first degree in x and y and therefore represents
a straight line ;
(b) is satisfied by the coordinates of the point of intersec
tion of the two given lines. Why ?
Example 1. Eind the equation of the line through the
point (2, — 5) and parallel to4:X{2y{5 = 0.
The system of lines parallel to 4:X \2y \ 5 = is given
by the equation y= — 2 x \7c. Since we want the particular
line of the system that passes through the point (2, — 5), the
equation must be satisfied by these coordinates. It follows
that, — 5 = — 4fA:orA;= — 1.
Therefore, y = —2x — 1 is the desired equation.
Example 2. Find the equation of the line through the
point (4, — 1) and perpendicular toSx\2y — 5 = 0.
The system of lines perpendicular to 3x\2y — 5 = is
given by the equation y = ^ x \ k. Since we want the line of
the system that passes through the point (4, — 1), we have
k = — y . Therefore, the desired equation is
y=2x_i_i or 2a; 31/ 11 = 0.
Example 3. Find the equation of the line through the
intersection of 2x\y — 4: = and x{y—l=0, and perpen
dicular to x\2y = S.
Any other line through the intersection of the given lines is
(4) (2x { y  4:)+ k (x + y  1) =0
92 MATHEMATICAL ANALYSIS [III, § 68
or
x{2 + k)\ y{l + k) + { i  k) = 0.
The slope of this line is — (2 + k)/{l + A:) and this must
be equal to the negative reciprocal of the slope of the straight
line X f 2 2/ = 3. Therefore,
2^^=2 and k=^.
1 + k 3
Substituting this value in equation (4) and simplifying, we
have 2x — y — S = 0, the required equation.
EXERCISES
1. Find the equation of the straight line through the point (1, 5) and
parallel to2x\Sy— 9=0; perpendicular to2x + 32/ — 9 = 0.
Alls. 2x +3?/17 = 0; 3x2y+ 7 =0.
2. Find the equations of the altitudes of the triangle whose vertices
are (2, 8), (4,  5), (3,  2).
3. Find the equation of the straight line through the intersection of
10x45^ + 11 = and x +2y + 14 = which is perpendicular to
x + 7y + l=0; parallel io S x  7 y = I.
4. Find the equation of the straight line through the intersection of
X \2y — 4: = and x — Sy^1=0 which is perpendicular to Sx — 2y
+ 4 = 0; parallel to x — y = 0.
5. Find the equation of the straight line through the intersection of
X \y  1 = 0, x3y+4 = and
(a) through the point (1, 1) ; Ans. x + 6y — 6 = 0.
(&) parallel to the line x + 2?/ — 9 = 0;
(c) perpendicular to the line 4 x — 5 ?/ = ;
((7) through the intersection of3x + 4?/ — 8 = and x— 5j/ + 7=0.
6. Find the equation of the straight line which passes through the
point
(a) (0, 0) and is parallel to2x — y + 4 = 0;
(6) (1, 2) and is perpendicular to3x— 2^/ — 1 =0;
(c) (— 1, 2) and is parallel to x — y — i=0.
7. Find the equation of the line which passes through the inter
section of X — 2^ + 2 = and x + y = and through the intersection of
x + y+2 = 0, x?/ = 0.
Ill, § 69]
LINEAR FUNCTIONS
93
8. Find the equation of the straight line through the intersection of
x2y {7 = and 2xyiS = and
(a) parallel to the ajaxis ;
(6) parallel to the «/axis.
9. Find the equation of the straight line which passes through the
intersection ofSx— ^ + 2 = and x + y = 6 and which
(rt) passes through the origin ;
(b) is parallel tox — 4?/ + 3=0;
(c) is perpendicular to 3aj — 2^ + 4 = 0.
69. Intersection of Two Lines. Simultaneous Equations.
We have just seen that linear equations in one or two vari
ables are represented in rectangular coordinates by straight
lines. We now wish to determine the coordinates of the point
of intersection of two lines whose equations are given. That
is, algebraically, we wish to find a set of values for x and y
which satisfy both equations.
Example 1. Solve the equations
(6)
(6)
Sx — 4:y = 7.
x\2y=9.
m
Multiplying equation (6) by 2 and add
ing the result to equation (5), we obtain
5 a; = 25, or X = 5.
Likewise multiplying equation (6 ) by 3 pj^ ^i
and subtracting the result from equation
(5), we have — 10 y = — 20, or y = 2. The set of values x = 5, y = 2 is
seen to satisfy both equations and is called the solution of the given
equations. If we plot lines (5) and (6) (Fig. 41), we see from their
graph that the coordinates of their point of intersection are (5, 2).
Therefore, a method of solving two linear equations in one or two
variables is to plot the lines represented by each equation, and then deter
mine from the graph the coordinates of the point of intersection. The
algebraic method of first eliminating one variable and then the other has
the advantage over the geometric method in that it is always accurate.
Instead of eliminating twice, the value found for either variable can be
substituted in either equation, and the value of the second variable de
termined.
94
MATHEMATICAL ANALYSIS
[III, § 69
Example 2.
Solve the equations
(7)
(8)
T
~
~
~
n

C\
^^
^
^A
^
•i
%
^
^'
^

?
gf*
^
^
y
^
n
Y"
■"
_
_
_
x2y = S.
X — 2 ?/ = — 5.
Subtracting the second equation from
the first, we obtain = 8. Tliat is, there
are no values of x and y satisfying both
equations. Such equations are said to
be inconsistent or incompatible. We see
that lines (7) and (8) have the same slope,
but different yintercepts, and therefore are parallel lines.
Example 3. Solve the equations
Fig. 42
(9)
(10)
xy = 2.
2x
2y=4.
Multiplying the first equation by 2 and subtracting the second from it,
we have 0=0. If equation (10) be divided by 2, equations (9) and (10)
are seen to represent the same relation between x and y, and are not
therefore sufficient to determine x and y. We can assign to either vari
able an arbitrary value and then find the corresponding value for the
other variable. The equations can, therefore, be said to have an infinite
number of solutions. Such equations are called dependent. The graphs
of these equations are coincident lines.
Let us now consider the general equations
(11) aiic + bill = Ci ,
(12) a^x^ b^ = C2,
where none of the constants are zero. Eliminating y, we
obtain {a^bz — a2&i) x = C162 — Cobi . Eliminating x, we ob
tain {oibo — aib^) y = aiC2 — a^c^ . Now if 0162 — 02^1 =^ ^) we
have
C261
x^"^^^^
aiC2 — «2Ci
we cannot
aib2 — 0.2^1 ^1^2 ~ ^2^1
If, however, aib^ — aa^i = 0, i.e. a.^ja^ = &2^i)
solve for x and y. Denoting the common value of these quo
tients by A:, we have ag = fcai , 62 = ^^1 • Then equations (11)
and (12) become a^x + b^y = Ci , and "ka^x f Icb^y = Co .
Ill, §69] LINEAR FUNCTIONS 95
We must now distinguish two cases according as C2 = kci or
C2 ^fc kci . In the former case, by dividing out k, we see that the
equations are dependent and have an infinite number of solu
tions. In the latter case they are inconsistent, and thus are
not satisfied simultaneously by any values of x and y.
Discuss the cases that arise if some of the constants are zero.
EXERCISES
Find, when they exist, the coordinates of the points of intersection of
the following lines. Check your answer from a graph.
1. 4a; + 2?/=:9. 3. x + 2y = 3. 5. a: + 4?/ = l.
2x5y = 0. 2x+4y = 6. 2x + 8?/ = 2.
2. 3x + 4?/ = 12. 4. x2y = 7. Z.x^y = 1.
X — y = b. 2x — iy = b, — x \2y = 3.
In the following exercises are the lines concurrent ? If so, what point
have they in common ?
7.x + 2?/ = 3. 8.xy = l. 9. x+2y = 5. 10.x2y = 3.
x — y = 0. 2x{y = S. 6x— y = 3. 5 x — y = 2.
5x— ?/=4. Sx2y = l. 2x + y=4. 2x + 3?/ = l.
In the following exercises, find k so that the lines shall be concurrent.
ll.x+y = 2. 12.2xy0. 13. 3 x  y = 4.
2x—y = l. x + 3y = 7. x + y = 0.
4x + y = k. Ans. b. x ^ ky = b. b x — 2 y = k.
14. The sides of a triangle have for their equations 2 x + y = 5,
a;— ?/ = 10, — 2x + ?/ = 6. What are the coordinates of the vertices of
this triangle ? What are the equations of the altitudes ?
15. Find the equation of the straight line through (2, 1), (—1, 2),
using the equation Ax + By + C = 0. [Hint : Solve for A/C and B/C]
16. Find the equation of the straight line through (4, 7) and having
the slope 3, using the equation Ax + By \ C = 0.
17. It has been shown experimentally, that the length I of a wire in feet
under a tension of p pounds, is I = a + bp, where a and b are constants.
Find a and bii 1= 190 when p = 270, and that I =190.2, when p = 450.
18. The readings T and 8 of two gas meters are connected by the
equation T = a } bS. Determine a and b when we know that 8 = 10,
when T = 300, and S = 100, when T= 420.
96 MATHEMATICAL ANALYSIS [III, § 69
19. The pull in pounds to lift a load I in pounds with a pulley is given
by the relation p =: al + b, where a and h are constants. Find a and b
when it is known that a pull of 8 pounds lifts a load of 40 pounds, while it
takes a pull of 2 pounds to hold the rope on when no weight is attached.
70. Equations Containing More than Two Unknowns. It
is easy to see that the methods employed in § 69 for solving
a system of two simultaneous equations, each containing two
unknown quantities, may also be employed for solving a
system of three or more equations, involving as many unknown
quantities as there are independent equations.
Example. Solve the equations
(13) 7x\Sy2z = 16.
(14) 5xy\ 6z = Sl.
(15) 2x + 6y + 3z = S9.
Adding three times (14) to (13) gives
(16) 22 X + 13 ;s = 109.
Adding five times (14) to (16) gives
(17) 27 a; + 28^= 194.
Solving equations (16) and (17) by the methods of § 69, we have a; = 2,
z = 5. Substituting these values in (13), we obtain y = i. It is readily
seen that x =2^ y = 4, z = b satisfies equations (13), (14), (15).
The cases in which three simultaneous equations in three
unknowns have no solution, or an infinite number of solutions,
will be discussed in Chapter XXI.
EXERCISES
Solve the following simultaneous equations :
l.>2x\4y{z = 12. 2. X + y + z = IS. 3. 2x Sy  z = 2.
^^x\yz = S. x2y{.iz = l0. 6xi2y + z = 8
x \ y + z = 7. 3x + y~Sz=5. x—2y — z = 2.
4. a; + 8 2/  4 = 9. 5. w + a 4 ?/ = 15. 6. a: + y = 4.
3x+3?/ — = 6. x + y {■ z = 19,. 2x\ z = 4.
bx + 2y2z = l. wj + ?/ + 0=17. y2 = 3.
10 + X + = 16.
Ill, §70]
LINEAR FUNCTIONS
97
7. If A and B can do a piece of work in 10 days, and A and C in 8
days, and B and C in 12 days, how long will it take each to do the work
alone ?
8. Three towns A, B, and C are situated at the vertices of a triangle.
The distance from A to B via C is 76 miles ; from A to C via B 79 miles;
from B to C via A 81 miles. Find the distance from A to B, from B to
C, from C to A.
9. In a triangular track meet the following was the final score :
SCOKE
First Place
Second
Place
Third Place
Total
College A . . .
College B . . .
College C . . .
5
2
2
3
4
2
3
1
6
37
23
22
How many points did each place count ?
10. Two passengers traveling from town A to town B have 500
pounds of baggage. The first pays $ 1.75 for excess above weight allowed,
the second $1.25. If the baggage belonged to the last passenger, he
would have to pay $ 4 excess. How much baggage is allowed to a
single passenger ?
11. A crew can row 4 miles downstream and back again in 1 hours,
or 6 miles downstream and half way back in the same time. What is
the rate of rowing in still water, and what is the rate of the current ?
Ans. 6 miles per hour ; 2 miles per hour.
12. Two trains are scheduled to leave two towns A and B, m miles
apart, at the same time, and to meet in h hours. The train leaving A
was k hours late in starting, so the trains met n hours later than the
scheduled time. What is the rate at which each train runs ?
13. Two men are running at uniform rates on a circular track 150 feet
in circumference. When they run in opposite directions, they meet every
5 seconds. When they run in the same direction, they are abreast every
25 seconds. What are their rates ?
14. Find a, &, c, so that y = a \ bx \ cx^ shall be satisfied by (2, 1),
(1,0), (3,6).
6a;2_ic3 a
16. Find a, 6, c, so that
x^
X + 1 X,
CHAPTER IV
THE QUADRATIC FUNCTION
I. GRAPHS OF QUADRATIC FUNCTIONS
71. The General Quadratic Polynomial ax^ h bx {c.
Having considered in some detail the linear function mx + b
and its geomietric interpretation, we now turn our attention to
a similar study of the quadratic function, i.e. sl function ex
pressed by a polynomial of the second degree in one variable.
Such polynomials are, for example, a;^ f 1, 100 f 50 ^ — 16.1 f^j
etc. The general form of such a polynomial is ax^ \ hx^ f c,
where a,h, c are constants and a^O. Such functions abound
in practice. Thus, if a projectile be shot vertically upward
from the top of a tower 100 ft. high, with an initial velocity of
50 ft. per second, the distance s (in feet) from the ground at
the end of t seconds, is given approximately by the poly
nomial
s = 100f 50i16.1f2.
The general formula for the distance s from the ground at the
end of t seconds of a projectile shot vertically upward is
(approximately)
where Sq is the distance from the ground when t = 0, Vq is the
initial velocity, and g is the socalled " gravitational constant,"
which varies slightly from place to place but is approximately
equal to 32.2 when the distance s is measured in feet and the
time is measured in seconds.
IV, § 72]
QUADRATIC FUNCTIONS
99
72. The Function x^. We consider first the simplest of all
quadratic functions, viz. the function y = x^. A brief tabular
representation of this function is as follows :
X
1
2
3
4
1
2
3
4
y
1
4
9
16
1
4
9
16
If we plot these points, we obtain Fig. 43, in which we notice
that the points seem to be arranged according to some regular
law. We may insert additional points by calculating values
~
Y
*
1
~\
^///A
t
'/, V'
z w,
wM
X
V
Fig. 43
of y for values of x between those already used. Thus
for X = 1.5, y = 2.25 and a; = — 1.5 y = 2.25. These points
are also marked on the figure. In general we see that for
X = a and also for a; = — a, we have y = a^. . Geometrically
this means that the graph is symmetrical with respect to the
2/axis, i.e. if the part of the graph on the right of the yaxis
is turned about the ?/axis until it falls in the original plane,
it will coincide with the part on the left of the ySixis. More
over, since x"^ is positive (or zero) for all real values of x, no
part of the graph will be below the a;axis.
100
MATHEMATICAL ANALYSIS
[IV, § 72
Keeping these facts in mind we shall make a more detailed
study of this function and its graph, by considering values
of X which are closer together. We shall confine ourselves to
values of x between x = and x = 2. The corresponding
values of y, for all values in this range at intervals of 0.1 of
a unit, are given in the following table :
X
y
X
y
X
y
X
y
0.1
0.01
0.6
0.36
1.1
1.21
1.6
2.56
0.2
0.04
0.7
0.49
1.2
1.44
1.7
2.89
0.3
0.09
0.8
0.64
1.3
1.69
1.8
3.24
0.4
0.10
0.9
0.81
1.4
1.96
1.9
3.61
0.5
0.25
1.0
1.00
1.5
2.25
2.0
4.00
We cannot, with any accuracy, insert in Fig. 43 the corre
sponding points of the graph. We therefore adopt a pro
cedure analogous to the use of a magnifying glass, in order to
separate the points. This we have done in Fig. 44 by choos
ing the unit on each axis 10 times as large as in Fig. 43. We
then see that there is no difficulty in plotting all the points
given in the above table.
Let us study more carefully the immediate neighborhood of
some point on the graph, for example, P(l, 1). We shall
magnify the shaded area in Fig. 44 in the ratio 10 : 1 and
make use of the following table :
X
y
X
y
X
y
X
y
X
y
0.90
.8100
0.95
.9025
1.00
1.0000
1.05
1.1025
1.10
1.2100
0.91
.8281
0.96
.9216
1.01
1.0201
1.06
1.1236
0.92
.8464
0.97
.9409
1.02
1.0404
1.07
1.1449
0.93
.8649
0.98
.9604
1.03
1.0609
1.08
1.1664
0.94
.8836
0.99
.9801
1.04
1.0816
1.09
1
1.1881
IV, § 72]
QUADRATIC FUNCTIONS
101
It will be noted that the points of the graph now lie almost
on a straight line (Fig. 45). We have drawn a straight line
through P for the purpose of comparison. If we should desire
a more detailed representation in the neighborhood of the
point P, we should calculate the values of y for values of x
between x — .99 and x = 1.01 and draw anew a small portion
w
1.2
1.1
1.0
0.9
i
1 2
Fig. 44
.90
1.0
Fig. 45
1.1
of the figure about P under a tenfold increase of the unit.
We would then find that the points would hardly be distin
guishable from the points on a straight line.
Similar conclusions might be reached near any other point
on the graph. It is of course impossible to prove this for
each separate point by separate calculations. To prove the
fact generally we proceed as follows. Let x^ be any particular
value of the variable x and y^ the corresponding value of the
function y ; then y^ = x^^. Now suppose that the value x
is increased or decreased by a certain amount, which we shall
call Ax (a decrease means that Ax is negative). The new
value of X is then Xi f Ax and the corresponding value of the
102
MATHEMATICAL ANALYSIS
[IV, § 72
function is {xi + Aa;)2, This new value of the function differs
from the original value of the function, yi, by a certain amount
which we shall call Ay. We then have
2/1 + A?/ = (xi h Axy
= Xi_^42xiAx + Ax"^',
but
2/1 = a^i
Therefore, by subtraction,
or
(1)
Ay = 2 cciAic + Ax'^
Ay = (2xi\ Ax) Ax.
Since formula (1) is true for every value of Xi , it follows that
Ay approaches zero when Ax approaches zero. This means
that in the neighborhood of the point {x^ , y^) we can find new
Fig. 46
points on the graph whose x and y differ from those of the
given point by as little as we please. This simply means that
the set of all points of the graph oi y = x^ form a set of points
with no gaps between them ; they form what we may call a
continuous line or curve.*
* A function is said to be continuous for a value x = Xi, if when Aa; ap
proaches the corresponding Ay also approaches 0. See footnote on p. 19.
IV, § 73] QUADRATIC FUNCTIONS 103
Further, equation (1) gives the relation,
^ = 2 iCi + Ax (if ^x 4 0).
Aa;
From the graph (Fig. 46) we clearly see that this change ratio
is the slope of the line joining the points P^ix^ , y^) and
P2(i>^i + Aa;, 2/1+ A?/).* If the latter point approaches the former
along the curve, i.e. if we let Ace become numerically smaller
and smaller, then the change ratio A?/ / Ao? will differ less and
less from 2xy. Indeed, we may choose Aic sufficiently small
(without making it zero) so that A?/ / Ax will differ from 2 x^
by less than any previously assigned amount.
Geometrically this means that in the immediate neighbor
hood of the point Pi on the graph oi y — a?^, the points of the
graph lie very near to the straight line through Pi whose slope
is 2 x^. From a somewhat different point of view, we can let
the secant joining the points Pi {x^ , y^) and Pg {xi\Ax, yi\Ay)
on the graph rotate about Pi in such a way that Aic, and there
fore Ay, become smaller and smaller and the secant approaches
a definite position. through Pi whose direction has the slope 2 Xi.
This line is by definition tangent to the graph at Pj , or the
graph is tangent to the line at Pi ; the point Pi is called the
point of contact. Combining the above results we have :
The graph of the function y = x"^ is a continuous curve, above
the Xaxis, symmetrical with respect to the yaxis, and passing
through the origin. At any point Pi {xi , i/i) on the curve, the
straight line ivith slope 2 Xi passing through this point is tangent
to the curve.
73. Further Observations regarding the Function y = x\
The preceding result tells us that when x= 1, the slope of the
tangent is 2. Reference to Fig. 45 will verify this result for
* This follows also directly from the formula m = {yz— 2/i)/(x« — ^i)
104
MATHEMATICAL ANALYSIS
[IV; § 73
the straight line there drawn, since this line has the slope 2.
In Fig. 47 we have reproduced Fig. 43 except that we have
replaced the several points plotted in the earlier figure by a
continuous curve and have drawn the tangent at the point
P(l, 1). Knowing that the slope of the tangent is 2, we can
easily construct the tangent. Starting from P we lay off any
convenient distance PM to the right and then lay off double
this distance MQ upward. The line PQ is then the required
tangent. A similar process leads to the construction of the
tangent at any other point of the curve.
From the fact that the slope of the tangent at any point on
the curve whose abscissa is Xi is 2 a^i, we see that as Xi increases
numerically the slope increases numerically, that is, the curve
becomes steeper and steeper the farther we go from the origin.
Also the slope is positive when x^ is positive and negative when
Xi is negative. This means that going from left to right the
curve slopes downward at the left of the origin, and upward at
the right of the origin. When a; = 0, the slope is zero, that is
to say, the tangent is parallel to the a;axis (here it coincides
with the ajaxis).
IV, § 73]
QUADRATIC FUNCTIONS
105
Hitherto in our drawings we have chosen the unit on the
t/axis to be equal to that on the ccaxis. This renders it im
possible to draw the graph of the function y = x"^ for large
values of a;, without making it of unwieldy size. However
t I
t 7
. _
 t 7
I } ^
^ ^ ^
t Z
r /
^z~
 z^i
// en
J U
t ^ ^
"I ::^^: : : :
A « i'
X ^
J ir
:: ~A~r : : :
^<^
: = r'V.. _ .
23456789 10
Fig. 48
nothing prevents us from choosing the unit on the t/axis
smaller than that on the icaxis, and in Fig. 48 we have chosen
it one tenth as large. A tabular representation is as follows :
X
±1
±2
±3
±4
±5
±6
±7
±8
±9
±10
y
1
4
9
16
25
36
49
64
81
100
In this case the slopes of the tangents are, respectively,
±2, ±4, ±6, ±8, ±10, ±12, ±14, ±16, ±18, ±20.
We have drawn the tangent at the point for which x = 5, and
have drawn the graph only for positive values of x.
Example. Find the equation of the tangent to the graph of
2/ = a;2 at the point (3, 9).
The slope of the tangent at the point (x^ , 2/1) is 2 a?] . There
fore at (3, 9) the slope is 6. The equation of the tangent is,
therefore, y — 9 = 6{x — S) oy y = 6x — 9.
106 MATHEMATICAL ANALYSIS [IV, § 73
EXERCISES
1. Discuss the functions y = — x^ ; y = 2x'^ \ y =— 2x^.
2. Construct for the point (2, 4) of the function y = x^ n. figure anal
ogous to Fig. 46. (Use a table of squares.)
3. Use the adjoining figure to give a geometric interpretation of the
X Ax equation Ay = 2 xiAx i Ax"^. The function y —x'^ is
Ax ^^^® interpreted as the area of the square whose side
is X.
f////////M
%
%
Ax
4. If in the function y = x^ we take x = 3, how
small must Ax be taken in order that Ay shall be
numerically less than 0.01 ? if we take x = 15 ? Is
the difference between these two results to be expected in view of the
nature of the graph ?
5. Draw the tangents to the curve 2/=x2 at the points for which x=0,
± i, ± 1, ± I, ± 2.
6. If X is the radius of a circle and y is its area, prove that the
change ratio Ay / Ax approaches the length of the circle as Ax approaches
zero.
7. Find the equations of the tangents to the curve ?/ = x^ at the
following points : (1, 1); (2,4); (1, 1); (2, 4). Construct the
tangents at these points.
8. The line perpendicular to the tangent at the point of contact is
called the normal to the curve at this point. Find the equations of the
normals to y = x^ at the points (1, 1) ; (2, 4) ; (— 1, 1) ; (— 2, 4). Con
struct each normal making use of its slope.
Ans. For the point (1,1): x + 2yS =0.*
9. Find the slope of the tangent to y = 3 x^ at the point whose abscissa
is xi. What is the value of this slope at the point (1, 3) ?
10. Find the equations of the tangent and the normal (see Ex. 8) to
y = 3x2 at the points (3, 27) ; ( 2, 12).
11. Find the points where the slope of the curve y = x^ has the values
 1 ; 2 ; 10.
12. 1 cu. ft. of water weighs 66.4 lb. What must be the diameter x
of a cylindrical can such that 1 in. of water contained in it will weigh
y oz. ? Plot the graph and find x when y = 50. Find y when x = 8.
* Assuming the units on the axis to be equal.
IV, § 75] QUADRATIC FUNCTIONS 107
74. The General Quadratic Function j/ = ax^ + 6x + c.
We may now dispose of the general case. Let
y = ax^ \ bx + c {a =^ 0)
be any quadratic function (in the case y = x^, a was 1, while
b and c were 0). Let x increase from the value Xi to the
value Xi f Ax, and suppose that this change in the value of x
changes the value of the function from 2/1 to yi f Ay. We desire
to calculate the value of Ay and of the change ratio Ay /Ax.
We have
2/1 f Ay = a(x^ + Axy\ b(xi + Ax) + c,
and
2/1 = a^i^ + b^i + c.
Subtracting, we obtain
(2) A?/ = (2axi\b \ a Ax) Ax,
and
(3) ^ = 2aXi\b{aAx (ifAa;^0).
Aa;
Equation (2) shows that Ay can be made numerically as small
as we please, by choosing Ax near enough to 0. Hence we may
say:
Every function of the form y = ax"^ \bx+ c is continuous.
Equation (3) shows that the change ratio Ay /Ax approaches
as a limit the value 2axi + bsiS Ax approaches 0. Hence we
may say:
The slope of the tangent to the curve y = ax^ \ bx { c at the
point whose abscissa is Xi is equal to 2aXi 4 b.
75. General Properties of the Function ax^ + 6x 4 c. The
discussion in the preceding section and the exercises have
furnished us with some information regarding some special
functions of the form aa^ {bx \ c.
108 MATHEMATICAL ANALYSIS [IV, § 75
It will now be shown that whenever the term in ar* is posi
tive {i.e. a is positive) the graph of the function is an inverted
a >o a < o
Fig. 49 Fig. 50
arch as in Fig. 49 and that whenever the term in x^ is negative
{i.e. a is negative) the graph is an arch like the one in Fig. 50.
To prove this we need only consider the slope of the tangent
to the curve as the point of contact moves along the curve.
We have just seen that the slope of the tangent is given by the
formula m — 2axi + & at the point whose abscissa is Xi . There
is just one point on the curve for which this slope is zero, viz.
the point whose abscissa is
^ a
Now let us write the slope m of the tangent in the form
The number in the parenthesis, i.e., Xi f 5/(2 a), is positive
when Xi> — 6/(2 a) and negative when Xi < — 6/(2 a). Geomet
rically this means that this parenthesis represents a positive
number for points to the right of the straight line x= —6/(2 a)
and a negative number for points to the left of this straight line.
Case 1 : a > 0. If a is positive, the slope m is positive for
points to the right of the line x = — 6/(2 a) and negative for
points to the left of this line.
In other words, for all points of the graph to the left of the
line x = — b/(2 a) the tangent slopes downward (as we go from
left to right) and for all points to the right of this line the
IV, § 76]
QUADRATIC FUNCTIONS
109
hi/>o
Fig. 51
tangent slopes upward. The point for which a: = — 6/(2 a) has
its tangent parallel to the ajaxis. This point is called the
minimum point of the graph (Fig. 51).
Case 2 : a < 0. Suppose on the other
hand that a is negative. The slope m
is then negative when Xi + &/(2 a) is
positive and positive when x^ + 6/(2 a) is
negative. The slope is therefore positive
when Xi< — 5/(2 a) and negative when
Xi> — b/(2 a). At the single point for
which X — — (6/2 a) the tangent is parallel
to the ajaxis. This point is called the maximum point of
the graph (Fig. 52).
When x = — 6/(2 a) the function y = ax^ \
bx \ c has a minimum value if a > and a
maximum value if a < 0.
The curve represented by the function
y = ax^ 4 6a; + c is symmetrical ivith respect
to the line x = —b/(2 a).
The proof is left as an exercise.
Hint. Show that the points which have abscissas — 6/(2 a) + h and
— 6/(2 a) — h have the same ordinate.
Fia. 52
76. Definitions.
of the form
The curve represented by an equation
y = ax"^ { bx { c
is called a parabola. The lowest (or highest) point on this
curve, i.e. the point for . which x = — 6/(2 a), is called the
vertex. The straight line through the vertex and per
pendicular to the' tangent at the vertex is called the axis
of the curve. The parabola is symmetrical with respect to
its axis.
110
MATHEMATICAL ANALYSIS
[IV, § 77
77. to draw the Graph of a Parabola y=ax'^\~bx^c.
The preceding discussion enables us to draw the graph of a
quadratic function without plotting many points.
Example 1. Sketch the graph of y = 2 x^ — 6 x \ 5.
The slope of the tangent at {x^, y^ is, by § 74:^ m = 4:Xi — 6.
The vertex of the curve is the point for which 4 ajj — 6 = 0, i.e.
the point for which x^ = 3/2 ; the corresponding value of y is
1/2 and the vertex is therefore the point (3/2, 1/2). This
point is the minimum point of the
curve. We plot this vertex V, draw
the horizontal tangent at this point
and the vertical axis. We desire a
few more points and their tangents
on each side of the axis and then
we can draw the curve. For ex
ample, we have
Fig. 53
X
y
m
1
1
2
2
1
2
5
6
Example 2. Sketch the graph
of 2/ = — aJ" + 4 a; f 5.
The slope of the tangent at
{xi , 2/i) is m = — 2 a^i f 4. The
vertex of the curve is at the point
for which — 2 a^i + 4 = 0, i.e. for
which Xi = 2. The corresponding
value of 2/i is 9. Therefore the
vertex, which is the maximum
point of the graph, is at (2, .9).
The graph is given in Fig. 54.
2 4 \ 6
Fig. 54
IV, § 77] QUADRATIC FUNCTIONS 111
EXERCISES
1. Tell which of the following functions have a maximum and which
have a minimum value. Find this value in each case and the correspond
ing value of X.
(a) 2x^ + Sx9.
Ans. Minimum value : — 17, when x =— 2.
(&) 3 x2 f 8 X  6.
(c) 6x^\l0x 12.
((?) 3 a;2 + 6 X  7.
(e) x^ + 1.
2. Find the coordinates of the vertex and the equation of the axis of
each of the following parabolas. Sketch the curves.
(a) 2/ = 2 ic2 + 5 X + 3.
(6) 2^ = 3 x2 + 9 X  6.
(c) y =  5 x2 + 10 X  12.
Ans. F= (1, — 7) ; axis, x=l.
(d) ?/ = 3 x2 + 6 X  7.
(e) y=x2+l.
3. The area of a certain rectangle in terms of the length of its side
X is ^ = X (100 — 2 x). Find x so that this area shall be a maximum.
4. A point moves on a straight line so that its distance s from a fixed
point O on the line at any time t is given by one of the equations below.
Draw the (s, t) graph and in each case show that the variable point
reaches, on one side of O, a maximum absolute distance from 0. Find
this maximum distance. Does this maximum absolute distance correspond
to a maximum or a minimum value of s ?
(a) s = f2_4« + 3. *
(6) s = 2fiSt + 10.
(c) s = 3 + 6 «  4 «2.
5. Find the equations of the tangent and the normal* to the curve
y = x2 — 3 X + 1 at the point (1,  1). Ans. y x; y = x2.
6. Find the equations of the tangent and the normal * to the curve
y = — 2 x2 + 3 X  1 at the point (1, 0) .
7. Find the equations of the tangent and the normal* to the curve
?/ = — 2 x2 + 4 X — 1 at the maximum point. Ans. y = I ; x = 1.
8. Find the equations of the tangent and the normal* to the curve
y = 3x2 — 6x+lat its vertex.
* See Ex. 8, p. 108.
112
MATHEMATICAL ANALYSIS
[IV, § 78
78. The Graph of y  k = a{x Kf. The face that the
graphs of functions of the form y = ax f 6a; 4 c, all have the
same general shajje but are differently located with respect
to the coordinate axes suggests that many of these graphs
may consist of curves, which might be brought into coin
cidence by a suitable motion in
the plane. That this is indeed
the case results from the follow
ing considerations, which lead to
a general principle of farreach
ing importance.
Suppose the graph of the equa
tion y = ax^ is moved parallel to
itself through a distance and
direction which carries the point to the point Q {h, k).
What will be the equation between the x and y of any point
P on the curve in its new position, the axes of coordinates
remaining in their original positon ? This question is readily
answered. Let P' be the position of P before it was moved.
The equation y — ax^ then tells us that M^P — a • OM'^ for
every position of P' on the curve in its old position. After the
motion, the directed segments OM' and 3PP' become respec
tively the (Erected segments QR and EP. Hence, for every
point P on the curve in its new position we have
(4) RP=a'QR\
If the coordinates of P are (x, y) we have x = Oil/, y = MP
and
QR = xh, RP=:yk.
Therefore, by (4), the curve in its new position is the graph
of the equation
(5) y — k = a ' (x — hy.
While we have applied these considerations to the function
IV, § 79] QUADRATIC FUNCTIONS 113
y = ax^, the reasoning is general ; consequently we may formu
late the following principle :
GEJfEEAL Principle. If in any equation between x and y ice
replace x by x — h and y by y — Jc, the graph of the neiu equation
is obtained from the graph of the origiyial equation by moving the
latter graph parallel to itself in such a ivay that the point moves
to the point (Ji, k).
We shall have occasion to apply this principle often in the
future.
79. Transformation by Completing the Square. At present
we may use the principle just stated to prove that the parabolas
y = ax"^ {bx^ c and y = ax^ are congruent curves.
This follows at once from the preceding general principle, if
we prove that the equation
(6) y = ax^ \bx\c
can be written in the form
(7) yJc= a{x  hy.
To do this we write (6) as follows :
2, = a(^ + ^+ )+o,
and then complete the square on the terms in the parentheses by
adding the term If' / (4 a^). In order to leave the value of y
unchanged we must also subtract a y. b^ / (4a2) = b^ /{4^a) from
the expression. This gives
rrr,^ f . ,bx , b^ \ , ¥
(7') y = a\x'^ + —' ' • ~
4^^;"^' 4a'
or
= a[x{
=)
^ ' 4a V 2
This is of the form (7) for the values
114 MATHEMATICAL ANALYSIS [IV, § 79
2 a 4tt
The operation just performed is called the transformation by
completing the square. It is found serviceable in a variety
of situations. It may be used to advantage in connection with
numerical examples.
Example. Discuss the graph of y = — 2 x^ + S x— 9.
We first write
y=2(a:24x+ )9;
and tiien
?/=2(:c24.x + 4)9 + 8
or
2/ + 1= 2 (a; 2)2.
The graph is then obtained from the graph ot y =—2x^ by moving the
latter parallel to itself so that its vertex moves to the point (2, —1).
EXERCISES
1. By reducing to the form y — k = a {x — h)^, discuss the graphs of
each of the following functions.
(a) y = 2x'^+12x\2. (d) y = 2x'^  1 x + S.
(6) 1/ =4x2 + 6a: 9. (e) y = Ax" + 1 x +2.
(c) ?/ =  3 x2 + 9 X + 10. (/) y =  3 x2  8 X + 10.
2. Show that the equation of the straight line y — yi = m (x — xi) may
be derived from the equation y = mx by the general principle of § 78.
3. The results of § 79 furnish a proof of the fact previously derived,
that the vertex of the parabola y = ax2 + 6x + c is at the point for which
x=— &/(2a). Explain.
4. Equation (7') proves that if a > 0, the value x = — 6 / (2 a) gives the
minimum value to y ; also that if a<0, the value x=— 6/(2a) gives
the maximum value to y. Explain without using the graph.
Write the following equations in the form a (x — hy + 6 (y — ^')2 = c,
where a, b, c, h, and k are constants.
6. x24x + 2?/28 2/ = 2. 8. x^ + y^iy = 2.
Ans. (x2)2 + 2(2/2)2 = 14. 9. a;28x + y2 = o.
'6. 2x2 + 4x+?/2_4y_3_0. 10. 3x24xy2 +2 =0.
7. 4 x2  4 X + 2 y2 _ 3 y ^ 1 _ 0.
IV, § 80]
QUADRATIC FUNCTIONS
115
II. APPLICATIONS OF QUADRATIC FUNCTIONS
80. Maxima and Minima. We have seen that a quadratic
function ax^ \bx + c has either a maximum or a minimum
value according as a is negative or positive. Numerous appli
cations involve the problem of finding this maximum or mini
mum value and the corresponding value of a;, as the following
examples show.
Example 1. A rectangular piece of land is to be fenced in and a
straight wall already built is available for one side of the rectangle.
What should be the dimensions of the rectangle in order that a given
amount of fencing will inclose the greatest area ?
Before beginning the solution proper we should note carefully the sig
nificance of the problem. The length of the fence being given, we may
use it to inclose rectangles of a variety of shapes, as indicated by the
dotted lines in Fig. 56. Some rectangle whose shape is between those
indicated will inclose the maximum area. .^,,^,,,,,,,,^^^^^^^^^
1
X
I — I
Fig. 50
To determine this shape is our problem.
To do this, it is necessary to express the  ' y j
area (the quantity we wish a maximum)
as a function of one variable.
Solution : Let the dimensions of the
rectangle be x and y and suppose the given length of fencing is L. We
then have
(8) 2x + y = L.
The area inclosed is J. = xy, which from (8) becomes
A = x(iL2x)=Lx2 x^.
Plotting this function, we have the parabola in Fig. 57. We desire to
find the value of x corresponding to the vertex
V of this parabola, for this gives the greatest
value to A. The slope m of the tangent is
given by the equation m = L — ix, and this
is zero (tangent horizontal at V) when x = \ L.
ZZ^ For this value oi x, y = ^ L. The maximum
area is therefore obtained when the width is
one half of the length. The maximum area
F.G. 57 is i L^ square units.
116
MATHEMATICAL ANALYSIS
[IV, § 80
Example 2. Three streets intersect so as to inclose a triangular lot
ABC. The frontage of the lot on BC is 180 ft. and the point A is 90 ft.
back of BC. A rectangular
building is to be constructed
on this lot so as to face BC.
What are the dimensions of
the ground plan which will
give the maximum floor
area?
In Fig. 58 we have drawn
the lot ABC and have indi
cated by dotted lines two
extreme plans. The ground
plan sought must be somewhere between these two extremes. To deter
mine its dimensions we proceed as follows :
Let X and y be the length of the sides of the ground plan. The floor
area (neglecting the thickness of the walls) is
(9)
A = xy.
In order to express A, for which we seek a maximum, in terms of x
alone, we now proceed to express y in terms of x. The triangles ABC
and ^ilifiV" are similar. (Why?) Hence we have
This gives
whence
(10)
From (9) and (10) we obtain
MN^^LA
BC DA
x 90
(Why ?)
180 90 '
y =  i a; H 90.
A = 90 X  I x^.
This expresses the floor .area as a function of the side x. The slope of
the tangent to the graph is given by
mz=90  X
and this slope is zero when x = 90, which in turn gives (by (10)) y = 45,
and therefore A = 4050. The maximum area is then 4050 sq. ft. and this
is obtained by making the building 90 ft. long and 45 ft. deep.
Draw the graph of the function A = 90x^ ^ x'^.
IV, §80] QUADRATIC FUNCTIONS 117
We may note that in both of these examples, the function of
which the maximum was sought was obtained as a function of
two variables. The conditions of the problem, however, made
it possible to express one of these variables in terms of the
other and thus to obtain the desired function as a quadratic
function of one variable, whereupon the solution was readily
effected. The difficulty in this type of problem is usually in
connection with the elimination of all but one of the variables.
To solve such a problem it is necessary to keep in mind the
following steps.
(1) Decide, and express in words, of what function a maxi
mum or a minimum value is to be found.
(2) Express this function algebraically.
(3) If this expression contains more than one variable, use
the conditions of the problem to find a relation or relations
connecting these variables.
(4) By means of the relation or relations found, eliminate
all but one of the variables from the function of which a maxi
mum or minimum value is sought.
(5) Proceed with the algebraic computation.
EXERCISES
1. The number 100 is separated into two parts such that the product
of the parts is a maximum. Find the parts and the corresponding
product. ^ns. 50, 50, 2500.
Is it possible to separate 100 into two parts such that the product of the
corresponding parts is a minimum ? Explain.
2. Prove that the rectangle of given perimeter which has the maxi
mum area is a square.
3. Find the greatest rectangular area that can be inclosed by 100 yd.
of fence.
4. Separate 20 into two parts such that the sum of their squares will
be a minimum.
118
MATHEMATICAL ANALYSIS
[IV, § 80
5. A man desires to build a shed against the
back of his house, the ground plan to be a rec
tangle. The roof is to be 1 ft. higher in the
back than in the front (see the adjoined figure).
He has on hand enough siding to cover 253 sq. ft.
Allowing 18 sq. ft. for a door and assuming that
the height from the ground to
the lowest part of the roof is 8
ft., what should be the dimen
sions of the ground plan in
order to get the greatest floor area ?
6. An underground conduit is to be built, the
cross section of which is to have the shape of a rec
tangle surmounted by a semicircle. If the cost of
the masonry is proportional to the perimeter, and if
the perimeter is 30 ft., what should be the dimensions of the cross section
in order that the conduit will have a maximum capacity ?
7. The same problem as in Ex. 6 with the perimeter of the cross section
given as a ft.
8. Determine the greatest rectangle that can be inscribed in a given
acute angled triangle whose base is 2 5 and whose altitude is 2 a.
*9. In the corner of a field bounded by two perpendicular roads a
spring is situated 8 chains from one road and 6 chains from the other.
How should a straight path be run by this spring and across the corner so
as to cut off as little of the field as possible ?
Ans. 12 and 16 chains from the corner.
81. Table of Squares. We have stated that the more
important functions have been tabulated (§ 28). The function
x^ is one of these. Tables of squares are very helpful in
shortening computation. A comparatively rapid method of
constructing such a table is given in Ex. 2 below. Here we
may make use of our knowledge of the function x^ to see that
for a sufhciently small interval in such a table, we are justified
in using linear interpolation (§ 56). Indeed we have seen that
* The function whose minimum is sought is not in this case quadratic
An approximate solution may be obtained graphically. The solution may be
computed by finding the slope of the graph from the definition of slope.
IV, §81] QUADRATIC FUNCTIONS 119
in a sufficiently small neighborhood of any point on the graph
of y = x^, the graph differs as little as we please from a straight
2 line. (See Fig. 45.) For example, if in the second
table on p. 100 we confine ourselves to only threeplace
.94 .884 accuracy, we find that the successive differences in
.95 .903 , p . , . , ,
g22 the function are almost proportional to the corre
.97 .941 sponding differences in the variable. We give in
.960 the adjoined table an extract from the table men
tioned. From this table we may conclude that
(.953)2 =.909.
This result is accurate only to the third decimal place.
EXERCISES
1. Find by interpolation from the above table the following :
(.954)2; (.981)2; (9.66)2; (9.89)2.
2. Compute by actual multiplication the squares of all the integers
from 31 to 40. This method of computing a table of squares becomes very
laborious. Write the results obtained from
your computation in a column, and write op
posite each pair of successive squares their
difference as shown in the adjoined beginning
of such a table. These differences are called
the flrst differences of the table. Do you ob
serve any regularity in the formation of these
differences? Prove in general the law here
suggested.
[Hint. Consider the difference between A;2 and (k + 1)^.]
Use this law to construct a table of squares from 41 to 100.
3. If the successive differences of the Jirst differences are formed, we
obtain the socalled second differences. Prove that in a table of squares
of successive integers the successive second differences are all equal to 2.
The first differences, therefore, have the character of a linear function.
Hence show how to compute the exact value of (.S2.6)2 from the value of
(32)2 and (33)2^ This process is known as quadratic interpolation.
X
a:2
Difference
31
32
33
961
1024
1089
63
65
34
35
36
120 MATHEMATICAL ANALYSIS [IV, § 82
III. QUADRATIC EQUATIONS
82. Definitions. An equation of the form ax^ \ bx + c = 0^
where a, b, and c are constants and a =^ 0, is called a quadratic
equation.
A value of x which when substituted in the equation
ax"^ \ bx { c = makes both members identical is called a root
Example 1. ^s 3 a root of the equation 2x^— 5x\6 = 0?
Substituting 3 for x, we find 2.3  5.3 +Q = 9 and not 0. Therefore
3 is not a root.
Example 2. Determine k so that one root of 2 kx — 3 ar + 5 = shall
be 1.
Since 1 is to be a root, we have 2 A: — 3  6 = 0, or k =— 1. The
e( [nation then becomes — 2 ic — 3 a; + 5 = 0.
83. The Roots of ax^^bx\c =0. It follows from § 79
that the equation ax^ f te + c = may be written in the form
5 \2 52 _ 4 ofc
V 2 ay
4a
provided a ^ 0. Dividing by a and solving for (x + 6/(2 a)),
we have ^ ^^_^
1^ + 2^>=^U?
ac
or
6 , V624ac
2a 2a
hence
(11)
b ± V624ac
2a
We have shown up to this point that if ax^ {bx\ c has the
value 0, then x must have one of the values given in equa
tion (11). We need still to prove the converse : If
— bHVb2 — 4flc — 6 — V62 — 4ac
X = X_r or x = ,
2a ' 2a
IV, § 82] QUADRATIC FUNCTIONS 121
then ax^ \hx\ c will have the value 0. This can be done by
substituting the values of x in turn in the expression ax^^hx^c
and simplifying the resulting expressions.''^
The last part of this proof is essential. We know that the
converse of a true theorem may be false. t The first part of
our discussion proved that no other values of x than those
given by (11) will satisfy the equation aoi^ \bx\ c = 0, but it
did not prove that either of these values does satisfy the given
equation.
Equations (11) maybe used as a formula for solving a quadratic
equation. Thus, solving
2 aj2 _ 5 aj _ 13 =
where a = 2, 6 = — 5, c = — 13, we have
or
5±V254(2)(13)
x_ 4
5 ± Vl29
Solution by Factoring. If the factors of a quadratic
equation may be found readily, one may proceed as in the
following example.
Example. Solve x"^  Sx \ 2 =0.
This equation may be written in the form
(a;2)(xl) = 0.
Therefore,
x2 =
or
X
1 = 0,
i.e.
x = 2
or
x=l.
Why? See § 48.
* The converse can be proved at present only if &2 _ 4 ac is not negative.
Why ?
t Thus the converse of the true statement, " A horse is an animal," would
be the false statement, " An animal is a horse."
122 MATHEMATICAL ANALYSIS [IV, § 83
EXERCISES
Determine whether the roots of the following equations are as stated.
1. x2_ 5a;46 =0 ;2, 3. 4. 2^2 _ 5^ + 3 = O; 1,  2.
2. a;2 + 5 x  6 = ; 1, 2. 5. a;2  7 = ; V7,  V7.
3. x2  12 a: + 30 = ; 5, 6. 6. 7 a:2 _ 2 x + 51 := ; 0, 1.
In the following equations determine k so that the number beside the
equation is a root. Find the other root.
7. a;2 + 2 Ax  5 = ; 1. 8. A:x2  6 x + A;2  1 = ; 0.
Ans. A; = 2 ; other root = — 5. 9. kx'^ — 6 A:x + 11 = fc ; 2.
Solve the following equations by means of the formula and also by
completing the square :
10. (ax + &)2 = 6 X. 15. sx2 + to  p = 0.
11. (x5)(7x3) = 12. jg x2 (3 x + 2)2 ^.^
12 y + 5 y^  5 ^ g . ■ 4 1
7 3 * 17. 3(5x2 10)+ 2x 5=0.
13. x2 f Ax  c?x2 + /i = 0. 18. x2 + (p g)xi)g = 0.
14. wP':r^ \ m{n—p)x — mp = 0.
Solve the following equations by factoring :
19. x2 _ 8 x + 15 = 0. 22. 3 x2  17 X + 10 = 0.
20. x2  14 X + 48 = 0. 23. • 5 X + 14 = x2.
21. 12  X  x2 = 0. 24. a&x2 + a2x + h'^x + a6 = 0.
25. A crosscountry squad ran 6 miles at a certain constant rate and
then returned at a rate 5 miles less per hour. They were 50 minutes
longer in returning than in going. At what rate did they run ?
Ans. 9 miles per hour.
26. When a single row of rivets is used to join together two boiler
plates, the distance p between the centers of the rivets is given by the
formula
^ = 0.56^+(?,
where t is the thickness of the plate and d is the diameter of a rivet hole in
inches. In a certain make of boiler the rivets are 1 inch apart and the
plate is \ inch thick. Find the diameter of the rivet holes.
27. How high is a box that is 6 ft. long, 2 ft. wide, and has a diagonal
8 ft. in length ?
IV, § 84]
QUADRATIC FUNCTIONS
123
28. The effective area JS" of a chimney is given by the formula
E = A — O.GVA^ Avhere A is the measured area. Find the measured
area when the effective area is 25 square feet.
29. Two men can row 12 miles downstream and back again in 5
hours. If the current is flowing at the rate of 1 mile per hour, how fast
can the men row in still water ?
30. Find the outer radius of a hollow spherical shell an inch thick
whose volume is 76 tt/S cubic inches.
[Hint. The volume of a sphere is 47rrY3.]
84. Graphic Solution. Example. Solve x^ — 4.x + 3=0
graphically.
Let us plot the graphs oi y == x^, y = 4:X— 3 with reference
to the same set of axes (Fig. 59). We see that the two graphs
intersect in two points, the coordinates of
which satisfy both equations. Therefore the
abscissas of these points are values of x
which make the righthand members equal,
i.e., for which
0^2 = 4a; 3
or
a;2_ 4a; 4.3^:0.
The roots are seen to be 1 and 3.
If the line and the parabola were tangent, what would you
say concerning the roots ? If the line and parabola do not
meet, what would you say concerning the roots ?
This problem may be solved graphically in
an entirely different way. We will plot the
curve y =zx^ — 4:X{3 (Fig. 60). The abscissas
of the points where this graph meets the a?axis
are roots of the original equation. Why ?
Describe the roots if the parabola touches
the icaxis. What would you say concerning the roots if the
parabola did not meet the icaxis ?
Fig. 59
Fig. 60
124 MATHEMATICAL ANALYSIS [IV, § 85
85. General Theorems. 1. If r is a root of the equation
ax"^ 4 6a; + c = 0, then x — r is a factor of ax"^ {hx \ c.
Dividing ax"^ + hx [ chy x — r, we obtain :
X — r\ ax^ \hx + c \ ax \{}j \ ar)
ax"^ — arx
{b h ar) X \ c
(b 4 ar) x — {b + ar) r
c { br \ ar"^'
Therefore
ax"^ \bx \ c= [cix + {b \ ar)']\x — r] + c + 5r + ar^.
But, by hypothesis, r is a root ; therefore, ar"^ + br + c = 0\
hence
ax^ \bx { G = \_ax \{b \ ar)'][_x — ?].
2. Prove that ifx — r is a factor of ax"^ \bx \ c, then x = r
is a root of ax^ \ bx + c = 0.
3. Prove that if the expression ax^ \ bx \ c is divided by
X — r, the remainder is ar^ \ br \ c.
The Discriminant of the Quadratic. In § 83 we saw
that the roots of the equation ax^ { bx \ c = are
6+V624ac ^^^ b V62_4(^c
The expression under the radical, namely, &2 _ 4 qc^ is called
the discriminant of the equation, because it enables us to dis
criminate as to the nature of the roots. From geometric con
siderations we know that a quadratic equation with real
coefficients a, b, c may have either two real distinct roots, two
real equal roots, or no real roots at all. The above formula
enables us to see the same truth algebraically.
If b^ — 4:ac = 0, we say that there are two real and equal
roots, each being — 6/2 a.
If 52 _ 4 fl(< > 0, there are two real and unequal roots.
IV, § 85] QUADRATIC FUNCTIONS 125
If 62 _ 4 ac < 0, there are no real roots. The roots of such
an equation are called imaginary or complex. The properties
of such numbers will be discussed fully in Chap. XVIII.
If the discriminant b^ — A ac is a perfect square and the
coefficients a, b, c are rational numbers, then the roots are
rational.
By finding the value of the discriminant we may determine
the nature of the roots of the quadratic without solving the
equation. Thus, in the equation Sx'^iix — 3 = 0, the dis
criminant is 52 and we conclude that the roots are real, un
equal and irrational.
Eelation of Roots to Coefficients. Let the roots of the
equation ax f 6a; f c = be denoted by /•, and ra . That is, let
— 64V62 — 4ac „ 1 —b — Vb"^ — 4 ac
and ?'2 =
2a 2a
By addition we have
5.V 52_4qe5V6'^4ac _ 2b_ b
'*^ + '*^ "" 2a ~ 2a~ a
By multiplication we have
_ r( b)^¥  4 ac\\{ b) + ^¥  4 ac']
'"'''' 4a^
_ 6^^ — 6^ 4 4 ac _ c
~ 4a2 a
Therefore, if we write the quadratic equation in the form
x^\^x + =0,
a a
the above results may be expressed as follows :
In a quadratic equation in ivhicJi the coefficient of the x^ term
is unity, (/) the sum of the roots is equal to the coefficient of x
with the sign changed; (ii) the product of the roots is equal to
the constant term.
126 MATHEMATICAL ANALYSIS [IV, § 85
EXERCISES
Solve graphically (two ways) each of the following equations :
1. 2x2 5x 3:^0. 3. 12a; = a;2. 5. 4x2 = 0.
2. a;28x + 15=0. 4. 2x23x5=0. 6. 4 + x2 = 0.
Form the equations with the following roots :
7. 4,5. Ans. x2+x20:=0. 9. 2+V5, 2V5.
8. V7,  V7. 10. c + 3 &, c  3 b.
11. What is the remainder when 3x2 — 2x + 5=0 jg divided by
x3? byx + 2? byx1? byx+1? [Hint : Use 3, § 85.]
Determine, the nature of the roots of the following equations :
12. 7x2_5x = 6. 14. 2 1/2 + 3 2/ + 24 = 0.
13. 2x = 73x2. 15. 9x2 = 4x5.
Determine k so that the following equations shall have equal roots.
[Hint : Place b'^ — iac equal to zero.]
16. kx^  6 x + 3 = 0. Ans. k=S. 18. x^ + 2 {1 ^ k)x { k^ = 0.
17. 3iK24^•x+ 2 = 0. 19. 2Arx2+(5A: + 2)x + 4^ + l=0.
20. Determine the limits on k so that equations 1619 shall have their
roots real and unequal ; imaginary and unequal.
21. If X is real, show that ~ must lie between and 1.
a;25x + 9 11
22. A party of students hired a coach for ^ 12, but three of the students
failed to contribute towards the expense, whereupon each of the others
had to pay 20 cents more. How many students were in the party ?
23. Cox's formula for the flow of water in a long horizontal pipe con
nected with the bottom of a reservoir is
Hd^ iv^ + ^^v2
' L 1200
where H is the depth of the water in the reservoir in feet, d the diameter
of the pipe in inches, L the length of the pipe in feet, and v the velocity
of the water in feet per second. If a reservoir contains 49 ft. of water,
find the velocity of the water in a 5inch pipe that is 1000 ft. long.
24. It takes two pipes 24 minutes to fill a certain reservoir. The larger
pipe can fill it in 20 minutes less time than the smaller. How long does
it take each pipe to fill the reservoir ? Ans. 60 min. ; 40 rain.
25. Prove algebraically and geometrically that if h^ — ^ac<:iQ^ the
value of the function ax^ + &x f c is positive for all (real) values of x,
if a > ; and negative for all (real) values of x, if a < 0.
IV, § 86] QUADRATIC FUNCTIONS 127
86. Equations involving Radicals. The method of solving
problems involving radicals will be illustrated by some
examples.
Example 1. Solve Va; + ^  2 (a; — 1)= 0.
Transposing the second term to the righthand member gives
V»T2 = 2(x 1).
Squaring,
x\2=^4.x'^Sx\4:, or 4a.'2 9 a; f 2 = 0.
Whence
, a; = 2, or i.
Do both these values satisfy the equation?
We have shown that, if VicH2 — 2 (ic — 1)=0, then x = 2
or J. But we cannot conclude conversely, that if a; = 2 or \,
then VxT2  2 (it  1) = 0.
In fact, if we substitute the values of x found in the original
equation, we find that a; = 2 is a root ; but a; = ^ is not.
Example 2. Solve the equation Va; + 8 + Vx + 3 = 5 Vx.
Squaring both sides, we find
a; f 8 + 2Va;2f llajf 24 + x + 3 = 25a;,
or
2 Va;2+ 11 a; + 24 = 23 x  11 ;
whence squaring, collecting terms, dividing by 25, we have
21 a;2  22 a; + 1 = ;
therefore, a; = 1 or ^V
What are the roots ?
EXERCISES
Solve each of the following equations :
1. Vic  2  3 = 0. 4.  \/4 X  3  Vx + 1 = 1.
2. >/x2(3 = 0. 5. v'x+5+ Vx+10=v2x + 15.
Ans. No roots. $. Vx4 b + Vx\ a = V2x+a + b.
3. Vx~+2VxTT =  1. 7. V2 X + 6  Vx + 4 = Vx  4.
Ans. 2. 8. Vx+3  V4 x + 1 = V2  8 x.
128 MATHEMATICAL ANALYSIS [IV, § 86
MISCELLANEOUS EXERCISES
Determine the condition existing among a, 6, c so that the equation
ax'^ + bx + c = shall have :
1. One root double the other. Ans. 2b^ = 9 ac.
2. The roots reciprocals of each other. Ans. a = c.
3. One root three times the other.
4. One root n times the other.
5. One root zero. Ans. c = 0.
6. One root equal to 1 ; 2 ; 3 ; n.
7. The roots numerically equal but opposite in sign. A7is. 6 = 0.
8. Find the area of the largest rectangle that can be inscribed in a
triangle whose base is 20 inches and whose altitude is 15 inches, if one
side of the rectangle is along the base of the triangle.
9. Separate twenty into two parts such that the product of half of one
part by a quarter of the other shall be a maximum.
10. Solve the equation y^ — Sy"^ ^ \b = 0. [Hint : Let y^ = x.]
11. Solve the equation fa; +  1 + T^ + ^1 — 12 = 0.
12. Solve the equation x^ + 8 x + 3 V^ + 8x42= 8.
13. Solve the equation — ^^^ = — •
x + l x^ 12
14. Find k so that the roots of {k + 2)x^ — 2 A;x + 1 = are equal.
15. Without solving, determine the sum and product of the roots of each
of the following equations :
(a) 2.r2_7x3 = 0. , (c) 4ic2 _ 3^: + 1 =0.
(6) x2  4 X + 2 = 0. (d) 2 a;2 + .3 x + 4 = 0.
16. Determine k so that the sum of the roots of the equation
2x2+(A l)x+(3A: 7) = 0is4. Ans. k = l.
17. Determine k so that the product of the roots of the equation
(2 A;  1) x2 + (A; + 3) » + (A;2  2 A; + 1) = is 2.
CHAPTER V
THE CUBIC FUNCTION. THE FUNCTION a^
87. The General Cubic Function ax^ f bx"^ \ ex \ d. Hav
ing discussed in the last chapter the general quadratic function
ax^ {bx \ c, we now turn our attention to
the general algebraic function of the third
degree, i.e. the general cubic function. It
is of the form
OQ^ \ bx^ \ ex + dj a=^ 0.
88. The Function ^. We begin with
the consideration of the function y = a^.
A brief tabular representation of this
function is given below.
We note that the values of x^
for negative values of x are the
same in absolute value as those
for the corresponding positive
values, but negative. If the
corresponding points are plotted
with respect to a pair of rec
tangular axes, we obtain Fig. 61.
The change Ay in y due to a
change Ax in x is calculated as follows, where x^ and y^ are
any pair of corresponding values of x and y :
(1) y^ + Ay = x^ + 3 a;i2 . Aa; H 3 x^^'^ + Aa^.
K 129
X
a;3
0.00
.5
0.12
1.0
1.00
1.5
3.36
2.0
8.00
2.5
15.62
3.0
27.00
^j^
x
in
\ ' '
3^'Alh
JrS^S^
Fig. 61
130
Since
this gives
(2)
MATHEMATICAL ANALYSIS
2/1 = ^ly
IV, § 88
A?/ = (3 x^^ + 3 x^^x + Aa;2) Aa;.
We can now conclude that as Aa; approaches zero, Ay also
approaches zero ; i.e. the function is continuous for all values
of X. From (2) we obtain
(3)
^ = 3a;i2j3a^iAa; + Aa2
Ao;
(if Ax z^ 0).
As Aaj (and, therefore, also A?/) approaches 0, this change ratio
approaches 3 x^. The slope m of the graph at the point {x^ , y^
is, therefore.
(4) m = ^x^\
This slope is positive for all values of Xy^
except a?! = 0. Why ? The function is
therefore an increasing function for all
values of x except a? = 0, i.e. at the origin,
where the graph of the function is tangent
to the Xaxis. The graph is exhibited in
Fig. 62, where we have drawn at certain
points the tangents to the graph by means
of (4) in order to insure greater accuracy.
 i^ :: J:::::
1 L .
± JL ::
"M : 3^:::::
di T
::^ : :t:::::
«' , i
'%
"r ' jL
Jrf T '1 e> 9 7: '.
:. ::::?_>:3 ^4i5
/ ^ _.
it >
"L — ":::: ::
:: : ■!«!: ::::: ::
Fig. 62
89. The Functions ax^ and a{x  hf + ft.
From the results of the last article and the
general principles previously established, we conclude that the
graph of the function
y = ay?
is obtained from that of 2/ = a^ by stretching or contracting all
the ordinates in the ratio  a  : 1, according as \a\ is greater
than 1 or less than 1, and in case a is negative reversing the
V, § 89]
CUBIC FUNCTIONS
131
signs of all the ordinates (Fig. 63).* Explain the reason for
this result.
The function y == a{x ~ Tif { k may be written in the form
y — 7c = a{x — hy.
Its graph is accordingly (§ 78) obtained from that of ?/ = aa^
by sliding the latter graph through a distance and in a direc
y=a(zh)3 + k
Fig. 64
tion represented by the motion from (0, 0) to (7 = {h, 7c). Ex
plain the reason for this (Fig. 64).
The slope of y = aa^ at the point (xi, y{) is 3 axi^. The slope
of a (a;  hy+ k at the point {x^, y{) is Sa{xi hy. The proof
of these statements is left as an exercise.
EXERCISES
1. From the graph of the function y — cc^, determine the volume of a
cube whose edge is 0.5 in. ; 0.5 ft. ; 3 f t. ; 1.5 cm.
2. Find the equation of the tangent and the normal to the curve y =.3^
at the point (2, 8) ; (_ 1, _ 1) ; (0, 0) ; ( 2,  8).
3. Draw each of the curves y= x^, y=4i x^, y=S(xiy, y=2(x\iy
* For example, if the unit on the yscale of Fig. 62 be doubled (i.e. made
equal to the a;scale) while the curve is left unaltered, the graph there given
will be the graph of y = i cc^.
132 MATHEMATICAL ANALYSIS [V, § 89
4. Show that the slope oiy = — ax^ at the point (xi, 2/1) is — 3 axi^.
6. Discuss the locus of y = — xK
6. Discuss the locus of y =— ax^ if a is positive and greater than 1 ;
less than 1. Show that the same curve will serve as the graph for all
values of a > if the units on the axes are properly chosen.
90. The Addition of a Term mx. Shearing Motion. If to
an expression in x defining a function, a term of the form mx
be added, the effect on the graph is readily described in terms
of a type of motion that is important in mechanics. For ex
ample, let us take the function a^ and investigate the effect
produced upon the graph by adding the
term —Sx. The graphs of y = a^ and
y=:—Sx are dravm in Fig. 65. The graph
of y = a^ — Sx is then obtained by adding
^ /" \A=r?5a the corresponding ordinates of the former
/ ^ graphs. The addition of these two func
tions is obtained graphically by sliding the
ordinate of each point on y = a^ vertically
up or down until the base of that ordinate
meets the graph of y =: — Sx. If we think of the ordinates
of 2/ = 25^ as attached to the avaxis and constrained to remain
vertical, the graph of y = a^ will become the graph of
y = a^ — 3x if the a>axis is rotated about the origin until it
coincides with the line y = — Sx. The resulting graph of
2/ = 0? — 3 a; is, of course, to be interpreted as drawn with ref
erence to the original a>axis. The motion just described,
whereby y = 0^ is transformed into y = a^ — Sxj is called a
shearing motion or a shear with respect to y = — Sx.
In general, if the term mx is added to aa^, the graph of the
function cta^ {mx is obtained by subjecting the graph of cue*
to a shear with respect to the line y = mx. .If a and m have
the same signs, the effect is in the direction of straightening
V, § 90]
CUBIC FUNCTIONS
133
the graph ; if a and m have different signs, the effect is in the
direction of emphasizing the curvature.
These effects can be produced by drawing the original figure
on the edges of a pack of cards, or on the edges of a book, and
then shifting the cards (or sheets of paper) as shown in Fig. 66.
EXERCISES
Draw the graph of the following functions, making use of the shear :
1. y = Sx'^ + x. 5. y=:ic8 + xl.
2. ?/ = x2 + X. B. y=x^ + x^2.
3. y=x^x. 7. y = x^l.
4. y =  2 x3 + 4 x. 8. y = x2  4 X.
9. Show that y = mx is the equation of the tangent to the curve
= x^ 4 mx at the origin.
134
MATHEMATICAL ANALYSIS
[V, § 91
91. The Functions a(x  hf + m{x  h)\ k and ax^ + bx^
{ cx + d. We have seen that the graph of
(6)
y = ax^ f mx
has one of the following forms (Fig. 67) according to the signs
of a and m.
If such a graph is subjected to a parallel motion which
Fig. 67
carries the origin to the point {h, k), the equation of the graph
in its new position is (§ 78)
(6) y — k = a{x — hy \m{x— h),
which when expanded takes the form
(7) 2/ = «^' — 3 ahx^ + (3 ah^ \m)x— ah^ — mh f k.
This is of the general form
(8) y =z ao? ^ hx^ [ ex + d.
V, § 93] CUBIC FUNCTIONS 135
Moreover, it includes for all values of 7i, k, m all the equations
of the general form (8). For (7) and (8) will be identical if
(9)  3 aA = 6, 3 a/i2 ( m = c,  ah^  mh { k = d.
The first of these equations determines h(a^O) ; h being
known, the second equation determines m; m and h being
known, the third equation determines k. ■ We may conclude
then that the graph of any function of the form (8) has one of
the shapes given in Fig. 67, but with the origin moved to a
point {h, k) given by the equations (9).
In order to draw the graph of a function of the form (8)
we could first transform (8) into the form (6) and then pro
ceed as in § 90. It is more expeditious, however, to proceed
more directly by making use of the slope of the function (8)
and our knowledge of what shapes may be expected.
92. The Slope of y = ax^ + bx^ \cx + d. The change Ay
in y due to a change Ax in the function
y = aa^ \ bx^ {ex j d,
when a; = aji , is
Ay = (3 axi^ { 2 bx^ \ c \ 3 ax^Ax + bAx f aAx'^) Ax.
This equation shows that the graph is continuous. Why?
When Ax approaches 0, the change ratio Ay /Ax approaches
the slope m, by definition. This gives,
93. To draw the Graph of y = aj^ h bx^ \ ex { d. We
shall illustrate by means of two examples the method of draw
ing the graph of a cubic function.
136
MATHEMATICAL ANALYSIS
[V, § 93
Example 1. Draw the graph oi y = a^ + x'^ — x + 2.
The slope m at the point (a;i, yi) is (§ 92)
m = S Xi^ \ 2 Xi — 1.
We seek first the points (if such exist) at which the tangent
is horizontal, i.e. where m = 0. The roots of the equation
m = 0, viz.
3a;i2 + 2a;] 1 =
are Xi = — 1 and Xi = . The slope is therefore at the
points (1,3) and (^, If).
We now compute a table of corresponding values of x, y, m
for values of x on both sides of and between x — 1 and x = ^.
Such a table and the corresponding figure are given below.
X
y
m
3
2
 13
20
7
1
3
2
1
1
If
3
4
2
12
15
Fia. 68
Example 2.* Draw the graph of y = — a^ — Sx^ — Sx^l.
The slope at the point where x = Xi is
(3a;i2 + 6aJi + 5).
m
Since the roots of the equation 3 iCi^ f 6 cci + 5 = are im
aginary, the graph has no horizontal tangents and the slope m
is negative at every point. We accordingly make a table of
values and construct the graph (Fig. 69).
V, § 94]
CUBIC FUNCTIONS
137
X
y
m
8
16
15
2
7
 5
1
4
 2
1
 5
1
8
14
Fia. 69
94. Maxima and Minima. We extend our definition of
maximum and minimum given in § 75 as follows :
A value of x for which a function stops increasing and
begins to decrease is said to correspond to a maximum of the
function ; a value of x for which the function stops decreasing
and begins to increase is said to correspond to a minimum of
the function. Thus in Ex. 1, § 93 the value x = —1 corre
sponds to the maximum 3 of the function ; the value a; = J
corresponds to the minimum ^ of the function.*
EXERCISES
Draw the following curves and locate in each case the maximum and
minimum points if there are any :
6. y = oc^ ■\ X + 1.
7. y = oc^_.
8. y = x^ — X.
9. y =z x^ + 2 x"^ h X.
1.
y = x^ + x^.
2.
v=f5!a..,
3.
y^X^ — 2x\\'
4.
y = x^x'^5x+2.
6. y = 2a;8 4^4x + l.
10. y
cfi x^ \x—l.
* Note that a maximum of a function does not mean the greatest value a
function can assume. In Ex. 1, § 93, the value of the function is greater when
x = 2 than when x = —l. It does mean a value of the function which is
greater than the values in the immediate neighborhood.
138
MATHEMATICAL ANALYSIS
[V, § 9;
95. Geometric Problems in Maxima and Minima. The
theory just explained has an important application in solving
problems in maxima and minima, i.e. the determination of the
largest or the smallest value a magnitude may have which
satisfies certain given conditions.
As we saw in § 80, the first step is to express the magnitude
in question algebraically. If the resulting expression contains
more than one variable, other conditions always will be given
which will be sufficient to express all of the variables in terms
of one of them. When the magnitude in question is expressed
in terms of one variable, we can proceed as in § 92 to find any
maximum or minimum values which there may be.
Example 1. Find the greatest cylinder that can be cut
from a given right circular cone, whose height is equal to the
diameter of its base.
Fig. 70
Let li be the given height of the cone and x and y the un
known dimensions of the cylinder (Fig. 70). Then the volume
V of the cylinder is equal to irx^y. But from similar triangles
we have
V, § 95] CUBIC FUNCTIONS 139
Therefore,
whence
Now
F= TTxXh 2x) = Trhx^  2 no^,
m = 2 irhx — 6 ttx^.
The roots of the equation m = are a; = and x =h/S.
It is left as an exercise to draw the graph of the function
V = Trhx"  2 TTCl^
and show that the value x = h/3 corresponds to the maximum
of the function, i.e. to y = h/S. Therefore the maximum
volume of the cylinder is obtained when the altitude is equal
to the radius of the base. The maximum volume is 7r^'/27
or 12/27 of the volume of the cone.
EXERCISES
1. A square piece of tin, the length of whose side is a, has a small
square cut from each corner and the sides are bent up to form a box.
Determine the side of the square cut away so that the box shall have the
maximum cubical contents, Ans. a/6.
2. Assuming that the strength of a beam with rectangular cross section
varies directly as the breadth and as the square of the depth, what are
the dimensions of the strongest beam that can be sawed from a round log
whose diameter is d. Ans. Depth = Vf d.
3. Find the right circular cylinder of greatest volume that can be in
scribed in a right circular cone of altitude h and base radius r.
Ans. Radius of the base of the cylinder equals f r.
4. Equal squares are cut from each corner of a rectangular piece of
tin 30 inches by 14 inches. Find the side of this square so that the re
maining piece of tin will form a box of maximum contents.
5. Show that the maximum and minimum points on the curve
y = x^ — ax + 6 (a > 0) are at equal distances from the yaxis.
6. Find the maximum volume of a right cone with a given slant
height L.
140
MATHEMATICAL ANALYSIS
[V, §96
96. The Power Function. The functions x" and l/a;», where
n is any positive integer, are called power functions of x. The
curves y = .t'» (Fig. 71) are known as parabolic, while the curves
y = l/iC* (Fig. 72) are known as hyperbolic.
The curves of the parabolic type possess the property that
they all pass through the point (0, 0) and the point (1, 1).
r
Y
T
H
"
\
4
. ili/L
\
*f
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4
^
\
a
>.
\
H
g
>%
.
'■
Tr
^
._, .
V
, V
/
/
\
u
>
8
C
\/
\
>
V
f
\
..
/
/
\
4) C*J
/
\
>'
/
N
ill
\
w
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4i
\
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m
s,
/
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fit
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s
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^
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A
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Z'
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PoLJv,
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i
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tsi
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orp
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Fig. 71
The larger the value of n, the greater is the slope of the tan
^ent at the point (1, 1).
The curves of the hyperbolic type all pass through the
point (1, 1). As X approaches 0, the corresponding value of y
becomes infinite. At a; = the value of y is undefined. As x
becomes infinite, the corresponding value oiy approaches 0.
V, § 96]
CUBIC FUNCTIONS
141
y
_1M.
1
Jv J]J
111
^
•< ^
~\h
,>^
^
il.
V^l
^
y
jLiiL
Hr
SpL
~^
j
^\ \ 11
T_^
VjLl
"^ ::ix
l_
/
J4l
1 1
1
\\
[7
\\
\
\t
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vl
1
^
J
^
V,
■^
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V'^
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f
rlO^
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■^&
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li
X y.
m^^
■^~'
=
^~'*p
CT
Y)

rrr
gmijj,.^]
s
f
1
1
"" "—
"L
^^\
_i_
:k
s\
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lyperbo
■ck
^'S
1
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1
1
r/ /»# L
x
.^i?:._
fo
"has
1,2,
tv/c
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\
r.
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J.
Fig. 72
EXERCISES
1. Draw the curves y =x^\ y = x^ ; y = x^; y = x^.
2. Draw the curves y = l/x; ?/ = l/x^ ; y = 1/x^ ; y = 1/ic*.
3. Prove that the slope of the tangent at the point (1, 1) to the curve
y = x2, is 2 ; to the carve y—x^ is 3 ; to the curve y=x^ is 4 ; lo the curve
y = x^ is 5.
4. Prove that for every even value of n, the parabolic curves ' ?/ = x"
pass through the point ( 1, 1); and that for every odd value of n, they
pass through the point (— 1, — 1).
6. Prove that the function x^ is an increasing function for all values
of x.
6. Find the equation of the tangent and the normal to y = ar^ at the
point (2, 32).
7. Prove that the slope of the curve y = 1/x at the point (xi, yi) Is
— 1/xi*. [The curve y — l/x is called a hyperbola.']
142
MATHEMATICAL ANALYSIS [V, § 96
ace* be obtained from the
8. How can the graph of the function y
graph of y = x" if a is positive ? negative ?
9. Find the equation of the tangent and the normal to the curve
y = l/a:2 at the point (2, \) .
10. Prove that all hyperbolic curves lie within the shaded regions of
the adjoining figure, while all parabolic curves lie in the regions left
unshaded.
CHAPTER VI
THE TRIGONOMETRIC FUNCTIONS
97. The functions we have discussed hitherto, namely, the
functions of the form mx ib, ax^ ^bx \ c, ax^ + bx"^ + ex \ dj
have all been defined by means of explicit algebraic expres
sions. They are all examples of a very large class of functions
known as algebraic functions. We now turn our attention to
functions defined in an entirely different way. As we shall
see, these functions depend on the size of an angle. They
enable us to express completely the relations between the
sides and the angles of a triangle, and they are of the
greatest practical importance in surveying, engineering, and
indeed in all branches of pure and applied mathematics.
98. Directed and General Angles. In elenJentary geometry
an angle is usually defined as the figure formed by two half
lines issuing from a point. However, it is often more serviceable
to think of an angle as being generated
by the rotation in a plane of a halfline
OP about the point as a pivot, start y^ \
ing from the initial position OA and / ^^^"'^ \
ending at the terminal position OB (Fig.
73). We then say that the line OP has
generated the angle AOB. Similarly, if OP rotates from the
initial position OB to the terminal position OA, then the angle
BOA is said to be generated. Considerations similar to those
regarding directed line segments (§ 6) lead us to regard one of
143
144 MATHEMATICAL ANALYSIS [VI, § 98
the above directions of rotation as positive and the other as
negative. It is of course quite immaterial which one of the
two rotations we regard as positive, but we shall assume from
now on, that counterclockwise rotation is
positive and clockwise rotation is negative.
Still another extension of the notion
of angle is desirable. In elementary
geometry no angle greater than 360°
is considered and seldom one greater than 180°. But from the
definition of an angle just given, we see that the revolving
line OP may make any number of complete revolutions before
coming to rest, and thus the angle generated may be of any
magnitude. Angles generated in this way abound in practice
and are known as angles of rotation.^
When the rotation generating an angle is to be indicated, it is
customary to mark the angle by means of an arrow starting at
the initial line and ending at the terminal line. Unless some
such device is used, confusion is liable to result. In Fig. 75
e^ 0^ Q
30' 390" 750' lilO
FiQ. 75
angles of 30°, 390°, 750°, 1110° are drawn. If the angles were
not marked one might take them all to be angles of 30°.
99. Measurement of Angles. For the present, angles will be
measured as in geometry, the degree (°) being the unit of measure. A
complete revolution is 360°. The other units in this system are the
minute ('), of which 60 make a degree, and the second ("), of which 60
make a minute. This system of units is of great antiquity, having been
* For example, the minute hand of a clock describes an angle of — 180®
in 30 minutes, an angle of — 540° in 90 minutes, and an angle of —720° in
120 minutes.
VI, § 101] TRIGONOMETRIC FUNCTIONS
145
used by the Babylonians,* The considerations of the previous article then
make it clear that any real number, positive or negative, may represent an
angle, the absolute value of the number representing the magnitude of
the angle, the sign representing the direction of rotation.
100. Angles in the Four Quadrants. Consider the angle
XOF = e, whose vertex O coincides with the origin O of a system of rec
tangular coordinates, and whose initial line OX coincides with the positive
p^Y. 0^^
Fig. 76
half of the xaxis (Fig. 76). The angle d is then said to be in the first,
second, third, or fourth quadrant, according as its terminal line OP is in
the first, second, third, or fourth quadrant.
101. Addition and Subtraction of Directed Angles. The
meaning to be attached to the sum of two directed angles is analogous to
that for the sum of two directed
line segments. Let a and h be
two halflines issuing from the
same point and let (a&) repre
sent an angle obtained by rotat
ing a halfline from the position
a to the position 6. Then if we
have two angles (a&) and (6c) with the same vertex 0, the sum (a6)(6c)
of the angles is the angle represented by the rotation of a halfline from
the position a to the position h and then rotating from the position 6 to the
position c. But these two rotations are together equivalent to a single rota
tion from a to c, no matter what the relative positions of a, &, c may have
* The terms minutes and seconds are derived from their Latin names, which
are partes minutss primse and partes minutx secundss. At present there is
a slight tendency among some authors to divide the degree decimally instead of
into minutes and seconds. Still other authors use the degree and minute and
divide the minutes decimally. Exercises involving both these systems will be
found in the text. When the metric system was introduced at the end of the
eighteenth century it was proposed to divide the right angle into 100 parts, called
grades. The grade was divided into 100 minutes and the minute into 100 sec
onds. This system is used in some European countries, but not at all in America.
146 MATHEMATICAL ANALYSIS [VI, § 101
been. Hence, we have for any three halflines a, 6, c issuing from a point 0,
(1) (a&) + (6c) = (ac), (ab) + (&a) = 0, (ab) = (cb)(ca).
The proof of the last relation is left as an exercise.
These relations are analogous to those of § 35 ; but an essential difference
must be noted. Given two points A and 5 on a line, we may speak of the
directed segment AB. The measure of AB is completely determined
when A and B and the unit of measure are given.
But if the halflines a and b are given, the angle
(ab) may be any angle generated by a rotation from
a to 6. Such angles may be positive or negative and
may involve, in addition to the minimum rotation
from a to &, any number of complete revolutions.
It is to be noted, however, that all possible determi
nations of the angle (ab) differ among themselves only by integral multi
ples of 360°. In other words, if 6 represents the smallest positive measure
(in degrees) of an angle from a to b, then any determination of (ab) is
given by the relation (ab) = 6 ± n 360° (n an integer). The equality
signs in relations (1) are then to be interpreted as meaning eqtial, except
for multiples o/360°.
If the position of the halfline h is determined by
the angle di which it makes with a given horizontal line
OX, and the position of another halfline h is deter
mined by the angle 62 which it makes with OX we have
angle from Zi to ^2 = 60 — 61 ,
except for multiples of 360°. Why ?
EXERCISES
1. What angle does the minute hand of a clock describe in 2 hours and
30 minutes ? in 4 hours and 20 minutes ?
2. Suppose that the dial of a clock is transparent so that it may be read
from both sides. Two persons stationed on opposite sides of the dial ob
serve the motion of the minute hand. In what respect will the angles de
scribed by the minute hand as seen by the two persons differ ?
3. In what quadrants are the following angles : 87° ? 135°?  326° ?
540°? 1500°? 270°?
4. In what quadrant is 6/2 if ^ is a positive angle less than 360° and in
the second quadrant ? third quadrant ? fourth quadrant ?
5. By means of a protractor construct 27°} 85° f (—30°) f 20° (— 45°).
6. By means of a protractor construct — 130° \ 56° — 24°.
VI, § 102] TRIGONOMETRIC FUNCTIONS
147
102. The Sine, Cosine, and Tangent of an Angle. We
may now define three of the functions referred to in § 97. To
this end let 9 = XOP (Fig. 80) be any directed angle, and let
O X ^
Fig. 80
w
T^
US establish a system of rectangular coordinates in the plane
of the angle such that the initial side OX of the angle is the
positive half of the i»axis, the vertex being at the origin and
the i/axis being in the usual position with respect to the
ajaxis. Let the units on the two axes be equal. Finally, let
P be any point other than on the terminal side of the angle
6, and let its coordinates^ be («, y). The directed segment
OP = r is called the distance of P and is always chosen posi
tive. The coordinates x and y are positive or negative accord
ing to the conventions previously adopted. We then define
ordinate of P y
The sine of 9
The cosine of d =
The tangent of Q =
distance of P
abscissa of P
distance of P
ordinate of P
provided x =^ 0,*
abscissa of P
These functions are usually written in the abbreviated forms
sin 0, cos 6, tan 6, respectively ; but they are read as " sine 0/^
" cosine 6/^ " tangent ^." It is very important to notice that
the values of these functions are independent of the position
of the point P on the terminal line. For let P{x', y') be any
other point on this line. Then from the similar right triangles
* Prove that x and y cannot be zero simultaneously.
148
MATHEMATICAL ANALYSIS
[VI, § 102
xyr* and a;'?/'r' it follows that the ratio of any two sides
of the triangle xyr is equal in magnitude and sign to the
ratio of the corresponding sides of the triangle xfy'r\ There
fore the values of the functions just defined depend merely
on the angle 6. They are onevalued functions of 6 and are
called trigonometric functions.^
Since the values of these functions are defined as the ratio
of two directed segments, they are abstract numbers. They
may be either positive, negative, or zero. Eemembering that r
is always positive, we may readily verify that the signs of the
three functions are given by the following table.
Quadrant
Sine
Cosine
Tangent
l'
+
+
+
2
+
3
+
4
103. Values of the Functions for 45°, 135°, 225°, 315°. In
each of these cases the triangle xyr is isosceles. Why?
Since the trigonometric functions are independent of the
position of the point P on the terminal line, we may choose
the legs of the right triangle xyr to be of length unity, which
^ j^l
P'
FiQ. 81
gives the distance OP as V2. Figure 81 shows the four angles
* Triangle xyz means the triangle whose sides are x,y,z.
t Trigonometric etymologically means relating to the measurement
of triangles. The connection of these functions with triangles will appear
presently.
VI, § 104] TRIGONOMETRIC FUNCTIONS
149
with, all lengths and directions marked. Therefore,
sin 45'=^,
V2
sin 135°
sin 225^
V2'
sin 315° =
V2'
vr
cos 45°= ■" ,
V2'
tan 45° = 1,
cos 135°= ^ ,
V2
tan 135°   1,
cos 225°= ^ ,
V2 .
tan 225° = 1,
cos315°=i:„
tan 315° =  1.
V2^
104. Values of the Functions for 30°, 150°, 210°, 330°. From
geometry we know that if one angle of a right triangle con
tains 30°, then the hypotenuse is double the shorter leg,
which is opposite the 30° angle. Hence if we choose the
shorter leg (ordinate) as 1, the hypotenuse (distance) is 2,
Jft' £n<^:^
v^
•VJ
.^^'
330
Fig. 82
and the other leg (abscissa) is V3. Figure 82 shows angles of
30°, 150°, 210°, 330° with all lengths and directions marked.
Hence we have
sin 30° =i
cos 30° = Y,
tan 30°=^,
V3
sin 150° =i
cos 150° = ^,
tan 150° = —,
V3
sin 210° = 
1
2'
cos 210°=^,
tan210°=J:^,
V3
sin 330°=
1
2'
cos 330° = ^^,
ton 330° = ^.
V3
150
MATHEMATICAL ANALYSIS
[VI, § 105
105. Values of the Functions for 60°, 120°, 240°, 300°. It is
left as an exercise to construct these angles and to prove that
sin 60° =
sin 120° =
sin 240° = 
sin 300° = 
106. Applications. The angle which a line from the eye to
an object makes with a horizontal line in the same vertical
plane is called an angle of elevation or an angle of depression.
V3
2 '
cos 60° = L
2
tan 60°= a/3,
V3
>
2
cos 120° = 1,
tan 120° ^  V3,
V3
2 '
COS 240°= ^,
tan 240° = v'3.
V3
2 '
cos 300° =,
2
tan300° = V3.
C^
Horizontal
^.^^^"'"""
Fia. 83
Horizontal
^^^
according as the object is above or below the eye of the observer
(Fig. 83). Such angles occur in many examples.
Example 1. A man wishing to know the distance between two points
A and B on opposite sides of a pond, locates a point C on the land (Fig.
84) such that ^C = 200 rd., angle G — Z^\ and angle B = 90°. Find the
distance AB.
^=sin(7. (Why?)
AG
AB=AC sin a
= 200 . sin 30°
= 200 . I = 100 rd.
Example 2. Two men stationed at points A and C 800 yd. apart and
in the same vertical plane with a balloon B, observe simultaneously the
angle of elevation of the balloon to be 30° and 45° respectively. Find the
height of the balloon.
Solution :
Fia. 84
VI, § 106] TRIGONOMETRIC FUNCTIONS
151
Solution : Denote the height of the balloon DB by y, and let DC = x
then AD = 800  x.
tan 45° = 1, we have 1 = ^
X
Since
and since tan 30° = 1/ V3, we have
V3 800  X
Therefore oc = y and 800 — x = y \' 8.
800
Solving these equations for y, we have
V3+ 1
292.8 yd.
EXERCISES
1. In what quadrants is the sine positive ? cosine negative ? tangent
positive ? cosine positive ? tangent negative ? sine negative ?
2. In what quadrant does an angle lie if
(a) its sine is positive and its cosine is negative ?
(6) its tangent is negative and its cosine is 'positive ?
(c) its sine is negative and its cosine is positive ?
(d) its cosine is positive and its tangent is positive ?
3. Which of the following is the greater and why : sin 49° or cos 49° ?
sin 35° or cos 35° ?
4. If 6 is situated between 0° and 360°, how many degrees are there in
6 if tan ^ = 1 ? Answer the similar question for sin = ^ ; tan ^ =  1. •
5. Does sin 60° = 2 • sin 30° ? Does tan 60° = 2 • tan 30° ? AYhat
can you say about the truth of the equality sin 2 ^ = 2 sin ^ ?
6. The Washington Monument is 555 ft. high. At a certain place in
the plane of its base, the angle of elevation of the top is 60°. How far is
that place from the foot and from the top of the tower ?
7. A boy \vhose eyes are 5 ft. from the ground stands 200 ft. from a
flagstaff. From his eyes, the angle of elevation of the top is 30°. How
high is the flagstaff ?
152 MATHEMATICAL ANALYSIS [VI, § 106
8. A tree 38 ft. high casts a shadow 88 ft. long. What is the angle
of elevation of the top of the tree as seen from the end of the shadow ?
How far is it from the end of the shadow to the top of the tree ?
9. From the top of a tower 100 ft. high, the angle of depression of
two stones, which are in a direction due east and in the plane of the base,
are 45° and 30° respectively. How far apart are the stones ?
^ns. 100(V3l)=73.2ft.
10. Find the area of the isosceles triangle in which the equal sides
10 inches in length Include an angle of 120°. Ans. 25 VS = 43.3 sq. in.
11. Is the formula sin2 ^ = 2 sin ^cos ^ true when = 30°? 60°?
120^^^^ ?
12. From a figure prove that sin 117° = cos 27°.
13. Find the tangent of the angle which the line joining the points
(^1) yi)i and (X2, Vi) makes with the ajaxis, assuming the units on the
two axes to be equal. Compare your answer with the definition of slope
in §§ 50 and 53.
14. Determine whether each of the following formulas is true when
e = 30°, 60°, 150°, 210° :
1 + tan2 d
1 +
cos'^ d
1 1
tan2 d sin2 0'
sin2 + cos2 = 1.
16. Let Pi(a:i, yi) and P2(a;2, 2/2) be any two points the distance be
tween which is r (the units on the axes being equal). If is the angle
that the line P1P2 makes with the ccaxis, prove that
X2X1 J y2  yi ^ 2 r.
cos 6 sin 6
107. Computation of the Value of One Trigonometric
Function from that of Another.
Example 1. Given that sin = f , find the
6y\ hs5 values of the other functions.
XB I _J__1^^ Since sin 6 is positive, it follows that d is
an angle in the first or in the second quad
rant. Moreover, since the value of the sine
is I, then y = Z  k and r = 6 •% where k is
any positive constant different from zero. (Why ?) It is, of course,
immaterial what positive value we assign to A;, so we shall assign the
Fia. 86
VI, § 108] TRIGONOMETRIC FUNCTIONS
153
value 1. We know, however, that the abscissa, ordinate, and distance
are connected by the relation x"^ + y^ = r^, and hence it follows that
X = ± 4. Fig. 86 is then selfexplanatory. Hence we have, for the first
quadrant, sin ^=f, cos^=, and tan^=; for the second quadrant,
sin ^ = , cos ^ = — I, tan 5 = — .
Example 2. Given that sin d = f^ and that tan d
is negative, find the other trigonometric functions of
the angle 6.
Since sin 6 is positive and tan 6 is negative, 6 must
be in the second quadrant. We can, therefore, con
struct the angle (Fig. 87), and we obtain sin ^ = j\,
:^1_
Fig. 87
108. Computation for Any Angle. Tables. The values of
the trigonometric functions of any angle may be computed by
the graphic method. For
example, let us find the
trigonometric functions of
35°. We first construct
on square ruled paper,
by means of a protractor,
an angle of 35° and choose
a point P on the ter
minal line so that OF
shall equal 100 units.
Then from the figure we
find that 0M= 82 units
iind MP = 57 units.
Fig. 88 Therefore
sin 35° = ^ = 0.57, cos 35° = ^\\ = 0.82, tan 35° =  = 0.70.
The tangent may be found more readily if we start by tak
ing OA — 100 units and then measure AB. In this case,
AB = 70 units and hence tan 35° = t% = ^•'^^•
It is at once evident that the graphic method, although
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154
MATHEMATICAL ANALYSIS [VI, § 108
10
Fig.
20 30 iO SO CO 70 60 90 100
89. — Graphical Table of Trigonometric Functions
VI, § 108] TRIGONOMETRIC FUNCTIONS 155
simple, gives only an approximate result. However, the values
of these functions have been computed accurately by methods
beyond the scope of this book. The results have been put in
tabular form and are known as tables of natural trigonometric
functions. These tables with an explanation of their use will
be found in any good set of mathematical tables.* In order
to solve several of the following exercises it is necessary to
make use of such tables.
Figure 89 makes it possible to read off the sine, cosine, or
tangent of any angle between 0° and 90° with a fair degree of
accuracy. The figure is selfexplanatory. Its use is illustrated
in some of the following exercises.
EXERCISES
Find the other trigonometric functions of the angle 6 when
1. tan^ = 3. 3. cos ^= if. 6. sin^ = f.
2. sin^=. 4. tan^=4. 6. cos^=^
7. sin ^ = I and cos 6 is negative. ■ _,
8. tan 6 = 2 and sin 6 is negative.
9. sin e = — ;^ and tan 6 is positive.
10. cos ^ = f and tan 6 is negative.
11. Can 0.6 and 0.8 be the sine and cosine, respectively, of one and
the same angle ? Can 0.5 and 0.9 ? Ans. Yes ; no.
12. Is there an angle whose sine is 2 ? Explain.
13. Determine graphically the functions of 20°, 38°, 70°, 110°. Check
your results by the tables of natural functions.
U. From Fig, 89, find values of the following :
sin 10°, cos 50°, tan 40°, sin 80°, tan 70°, cos 32°, tan 14°, sin 14°.
15. A tower stands on the shore of a river 200 ft. wide. The angle of
elevation of the top of the tower from the point on the other shore exactly
opposite to the tower is such that its sine is . Find the height of the
tower.
* See, for example, The Macmillan Tables, which will be referred to
in connection with this book.
156 MATHEMATICAL ANALYSIS [VI, § 108
16. From a ship's masthead 160 feet above the water the angle of de
pression of a boat is such that the tangent of this angle* is ^j. Find tlie
distance from the boat to the ship. Ans. 640 yards.
17. A certain railroad rises 6 inches for every 10 feet of track. What
angle does the track make with the horizontal ?
18. On opposite shores of a lake are two flagstaffs A and B. Per
pendicular to the line AB and along one shore, a line BC = 1200 ft. is
measured. The angle ACB is observed to be 40° 20'. Find the distance
between the two flagstaffs.
19. The angle of ascent of a road is 8°. If a man walks a mile up the
road, how many feet has he risen ?
20. How far from the foot of a tower 150 feet high must an observer,
6 ft. high, stand so that the angle of elevation of its top may be 23°. 5 ?
21. From the top of a tower the angle of depression of a stone in the
plane of the base is 40° 20'. What is the angle of depression of the stone
from a point halfway down the tower ?
22. The altitude of an isosceles triangle is 24 feet and each of the equal
angles contain 40° 20'. Find the lengths of the sides and area of the
triangle.
23. A flagstaff 21 feet high stands on the top of a cliff. From a point
on the level with the base of the cliff, the angles of elevation of the top
and bottom of the flagstaff are observed. Denoting these angles by a
and /3 respectively, find the height of the cliff in case sin a ■=. ^ and
cos/3 = H. Ans. 76 feet.
24. A man wishes to find the height of a tower CB which stands on a
horizontal plane. From a point A on this plane he finds the angle of ele
vation of the top to be such that sin CAB = f . From a point A' which
is on the line AC and 100 feet nearer the tower, he finds the angle of
elevation of the top to be such that tan CA'B — f . Find the height of the
. tower.
25. Find the radius of the inscribed and circumscribed circle of a regu
lar pentagon whose side is 14 feet.
26. If a chord of a circle is two thirds of the radius, how large an
angle at the center does the chord subtend ?
27. A boy standing a feet behind and opposite the middle of a football
goal observes the angle of elevation of the nearer crossbar to be a, and
the angle of elevation of the farther crossbar to be p. Prove that the
length of the field is a [tan a — tan /3]/tan /3.
VI, § 109] TRIGONOMETRIC FUNCTIONS 157
109. The Sine Function. Let us trace in a general way the
variation of the function sin ^ as ^ increases from 0° to 360°.
For this purpose it will be convenient to think of the distance
r as constant, from which it follows that
the locus of P is a circle. When = 0°, the
point P lies on the xaxis and hence the
ordinate is 0, i.e. sin 0° = 0/r = 0. As ^
increases to 90°, the ordinate increases
until 90° is reached, when it becomes equal
to r. Therefore, sin 90° = r/r = 1. As ^ p^^ ^
increases from 90° to 180°, the ordinate de
creases until 180° is reached, when it becomes 0. Therefore
sin 180° = 0/r = 0. . As ^ increases from 180° to 270°, the ordi
nate of P continually decreases algebraically and reaches its
smallest algebraic value when = 270°. In this position the
ordinate is — r and sin 270° = — r/r = — 1. When enters
the fourth quadrant, the ordinate of P increases (algebraically)
until the angle reaches 360°, when the ordinate becomes 0.
Hence, sin 360° = 0. It then appears that :
as 6 increases from 0° to 90°, sin increases from to 1 ;
as 6 increases from 90° to 180°, sin decreases from 1 to ;
as 6 increases from 180° to 270°, sin 6 decreases from to — 1 ;
as 6 increases from 270° to 360°, sin $ increases from — 1 to 0.
It is evident that the function sin 6 repeats its values in the
same order no matter how many times the point P moves
around the circle. We express this fact by saying that the
function sin 6 is periodic and has a period of 360°. In symbols
this is expressed by the equation
sin [6 + n . 360°] = sin 9,
where n is any positive or negative integer.
The variation of the function sin 6 is well shown by its
158
MATHEMATICAL ANALYSIS
[VI, § 109
graph. To construct this graph proceed as follows : Take a
system of rectangular axes and construct a circle of unit radius
Fig. 91
with its center on the ajaxis (Fig. 91). Let angle XOP = 0.
Then the values of sin 6 for certain values of 6 are shown in
the unit circle as the ordinates of the end of the radius drawn
at an angle 0.
d
30°
45°
60°
90°
...
sin^
M,Pi
M2P2
MsPs
M,P,
Now let the number of degrees in ^ be represented by dis
tances measured along OX. At a distance that represents 30°
erect a perpendicular equal in length to sin 30° ; at a distance
that represents 60° erect one equal in length to sin 60°, etc.
Through the points 0, Pi, P^," draw a smooth curve'; this
curve is the graph of the function sin 6.
If from any point P on this graph a perpendicular PQ is
drawn to the a;axis, then QP represents the sine of the angle
represented by the segment OQ.
Since the function is periodic, the complete graph extends
indefinitely in both directions from the origin (Fig. 92).
Y
VI, § 110] TRIGONOMETRIC FUNCTIONS
159
110. The Cosine Function. By arguments similar to those
used in the case of the sine function we may show that :
as 6 increases from 0°to 90°, the cos decreases from 1 to ;
as $ increases from 90° to 180°, the cos 9 decreases from to — 1 ;
as increases from 180° to 270°, the cos 6 increases from — 1 to ;
as increases from 270° to 360°, the cos increases from to 1.
The graph of the" function is readily constructed by a method
Fig. 93
similar to that used in case of the sine function. This is
illustrated in Fig. 93.
The complete graph of the cosine function, like that of the
sine function, will extend indefinitely from the origin in both
7
^^ /^
\ AA
Vy ''
\ / \^
i
^=cosx
Fig. 94
directions (Fig. 94). Moreover cos 0, like sin 0, is periodic and
has a period of 360°, i.e.
cos [6 + n . 360°] = cos e,
where n is any positive or negative integer.
160
MATHEMATICAL ANALYSIS
[VI, § 111
Fig. 95
111. The Tangent Function. In order to trace the varia
tion of the tangent function, consider a circle of unit radius
with its center at the origin of a system of rectangular axes
(Fig. 95). Then construct the tangent to
this circle at the point M{1, 0) and let P
denote any point on this tangent line. If
angle MOP = 0, we have tan 6 = MP/OM
= MP /I — MP, i.e. the line MP represents
tana
Now when 6 = 0°, MP is 0, i.e. tan 0° is 0.
As the angle 6 increases, tan 6 increases. As
6 ap]3roaches 90° as a limit, MP becomes
infinite, i.e. tan 6 becomes larger than any number whatever.
At 90° the tangent is undefined. It is sometimes convenient
to express this fact by writing
tan 90° = 00.
However we must remember that this is not a definition for
tan 90°, for oo is not a number. This is merely a short way of
saying that as $ approaches 90°, tan becomes infinite and
that at 90° tan 6 is undefined. See § 36.
Thus far we have assumed 6 to be an
acute angle approaching 90° as a limit.
Now let us start with 6 as an obtuse angle
and let it decrease towards 90° as a limit.
In Fig. 96 the line MP' (which is here
negative in direction) represents tan 9.
Arguing precisely as we did before, it is
seen that as the angle 6 approaches 90°
as a limit, tan 6 again increases in magnitude beyond all
bounds, i.e. becomes infinite, remaining, however, always
negative.
Fig. 96
VI, § 111]
TRIGONOMETRIC FUNCTIONS
161
We then have the following results. ,
(1) When 6 is acute and increases toward' 90° as a limit,
tan always remains positive but becomes infinite. At 90°
tan is undefined.
(2) When 6 is obtuse and decreases towards 90° as a limit,
tan 6 always remains negative but becomes infinite. At 90°
tan 6 is undefined.
It is left as an exercise to finish tracing the variation of the
tangent function as 6 varies from 90° to 360°. Note that
tan 270°, like tan 90°, is undefined. In fact tan n • 90° is unde
fined, if n is any odd integer.
360° X
Fia. 97
To construct the graph of the function tan 9 we proceed
along lines similar to those used in constructing the graph of
sin e and cos 0. The following table together with Fig. 97
illustrates the method.
d
0°
30°
45°
60°
90°
120°
136°
150°
180°
210°
tan^
MPi
MP2
MPz
undefined
MPi
MP^
MP^
3/P7=0
MPi
162 MATHEMATICAL ANALYSIS [VI, § 111
It is important to notice that tan 6y like sin 9 and cos 6j is
periodic, but its period is 180°. That is
tan (6 + 1 • i8o°)= tan 6,
where n is any positive or negative integer.
EXERCISES
1. What is meant by the period of a trigonometric function ?
2. What is the period of sin 6 ? cos 6 ? tan d ?
3. Is sin d defined for all angles ? cos d ?
4. Explain why tan 6 is undefined for certain angles. Name four
angles for which it is undefined. Are there any others ?
5. Is sin (^ + 360^) = sin d ?
6. Is sin {e + 180°) =sin^?
7. Is tan {d + 180°) = tan ^ ?
8. Is tan {6 + 360°) = tan ?
Draw the graphs of the following functions and explain how from the
graph you can tell the period of the function :
9. sin^. 11. tan^. 13. ^•
cos d
10. cosd. 12. — — 14. ^ .
sin d tan 6
Verify the following statements :
16. sin 90° + sin 270° = 0. 18. cos 180° + sin 180° =  1.
16. cos 90° + sin 0° = 0. 19. tan 360° + cos 360° = 1.
17. tan 180° + cos 180°= 1. 20. cos90°f tan 180°sin270° = l.
21. Draw the graphs of the functions sin ^, cos 6, tan ^, making use of
a table of natural functions. See p. 638.
22. Draw the curves y = 2 sin ^ ; y = 2 cos ; y = 2 tan 6.
23. Draw the curve j/ = sin + cos 6.
24. From the graphs determine values of d for which sin ^ = ^ ; sin 6
= 1 ; tan <? = 1 ; cos = ^ ; cos ^ = 1.
VI, § 112] TRIGONOMETRIC FUNCTIONS
163
Fig. 98
112. Polar Coordinates. It is convenient at this point to
introduce a new way of locating the position of a point in a
plane, and of representing the graph of a function. To this end
(Fig, 98) let OA be a directed line in the plane which we shall
call the initial line or the polar axis.
This line is usually drawn horizontally
and directed to the right. The point
is called the pole or the origin. Let P
be any point in the plane and draw the
line OP. The position of P is then
located completely if we know the angle ^OP=^and the dis
tance OP=:p. The two numbers (p, 9), called respectively the
radius vector and the vectorial angle, are known as the polar
coordinates of the point P.
In Fig. 98 we have represented a case in which and f> are
both positive. Either ot p or both may be negative under
the following conventions. The angle is positive or negative
according to the direction of its rotation, as in § 98. The
positive direction on OP is the direction from along the
terminal side of the angle 0, i.e., it is the direction into which
OA is rotated by a rotation through the angle 0.
With these conventions a point P whose polar coordinates
(/), 0) are given is completely de
termined. Figure 99 shows points
whose polar coordinates are (2, 30°),
(2, 30°), (2, 30°), and (2,
— 30°). It will be noted that, if p is
positive, P is on the terminal side of
$, while if p is negative, P is on the
terminal side produced through 0.
On the other hand, a given point P has an unlimited number of
polar coordinates (p, 6). Even if we confine ourselves to angles
is,so')
{2,30')
(s,so )
i2y30°)
Fig. 99
164 MATHEMATICAL ANALYSIS [VI, § 112
in absolute value less than 360°, a point lias in general /owr dif
ferent sets of polar coordinates. Fig. 100 shows that the same
(e.so').
point P may be designated by any one of the pairs of values
(2, 30°), (2,  330°), ( 2, 210°), and ( 2,  150°).
EXERCISES
1. Locate the points whose polar coordinates have the following values :
(4, 30°), (2, 45<^), (3, 60°), (2, 160°), (3, 90°), (2, 180°),
(2, 0°), (0, 90°), (2, 180°), ( 3, 270°).
2. For each of the points in Ex. 1, give all other sets of polar coordi
nates for which 6 is in absolute value less than 360°.
3. What exceptions are there to the statement " 6 being confined to
angles in absolute value less than 360°, every point has four and only
four distinct sets of polar coordinates " ?
4. Where are all the points for which ^ is a given constant ?
5. Where are all the points for which p is a given constant ?
113. Graphs in Polar Coordinates. Polar coordinates may
be used to represent the graph of a given function, in a way
quite similar to that in the case of rectangular coordinates.
Fig. 101 gives an example in which the idea of polar coor
dinates is used in practice. In this example the ^scale rep
resents time, the pscale represents temper ature.*^ Some forms
of selfrecording hygrometers employ the same idea.
* It will be noted that in this example the radius vector is measured along
a circular arc instead of along a straight line. This is due to the mechanical
(construction of the instrument. Cf. footnote, p. 9. The fundamental idea is,
nevertheless, that of polar coordinates.
VI, § 1131 TRIGONOMETRIC FUNCTIONS
165 '
In plotting the graph of a function in polar coordinates we
proceed as in the case of rectangular coordinates. A table of
Fig. 101
corresponding values of the variable 6 and the function p is
166
MATHEMATICAL ANALYSIS
[VI, § 113
constructed. Each such pair of values is then plotted as a
point, and a curve drawn through these points.
Example. Plot in polar coordinates the graph of p = sin 6. We ob
tain the table below. Figure 102 exhibits the corresponding points, with
(1.90)
e
p = sin ^
0°
.00
30^
.50
46°
.71
60°
.87
90°
1.00
120°
.87
135°
.71
150°
.50
180°
.00
210°
 .50
225°
 .71
240°
 .87
270°
 1.00
300°
 .87
315°
 .71
330°
 .50
360°
.00
a curve drawn through them. Observe that each point serves to represent
two pairs of corresponding values. Thus the pairs (^, 30°) and ( ~ i, 210°)
are represented by the same point. This curve suggests a circle, of diame
ter unity, tangent to the polar axis at the origin.
114. The Graph of sin 6 and cos 6 in Polar Coordinates.
We may now prove :
The graph, in polar coordinates, of the function p — sin $ is a
circle of diameter unity, tangent to the polar axis at the origin.
Let P (p, 6) be any point on such a circle (Fig. 103). Then,
for any value 6 in the first quadrant
OA 1
or
p = sin ^.
VI, § 114] TRIGONOMETRIC FUNCTIONS
167
Conversely, if p = sin 6, the point P is on the circle. Why ?
A similar proof, which is left as an exercise, may be given
when 6 is in the second, third, or fourth quadrants (Fig, 104).
Similarly, we may prove :
FiQ. 1(M
Hie graph of
p = GO8
in polar coordinates is a circle of diameter unity, passing through
the pole and having its center on the polar axis.
The proof of this statement is left as an exercise. See Figs.
105, 106.
On account of their simplicity, the polar graphs of sin $ and
cos 6 are very serviceable. It is for this reason that we have
FiQ. 105
Fig. 106
introduced them at this point. Polar coordinates will be dis
cussed again, particularly in Chapter XIV, and incidentally
in other chapters.
168 MATHEMATICAL ANALYSIS [VI, § 114
EXERCISES
1. From Fig. 101, find the temperature at 9 p.m. on Tuesday ; at 3 p.m.
on Monday. When was the temperature a maximum ? a minimum ?
2. Plot in polar coordinates the graph representing the variation in
temperature given in Ex. 1, p. 16.
3. Plot the graph in polar coordinates of the function p = tan d. Why
is this graph not convenient to represent the function tan 6 ?
4. Prove that the graph, in polar coordinates, of /> = a cos ^ is a circle
of diameter a, passing through the origin and w^ith its center on the polar
axis.
5. Prove a theorem regarding the graph of p = a sin d,
115. Other Trigonometric Functions. The reciprocals of
the sine, the cosine, and the tangent of any angle are called,
respectively, the cosecant, the secant, and the cotangent of
that angle. Thus,
cosecant = distance of P ^ r ( i^^^ ^ o).
ordinate of P y ^^ ^ ^
, /J distance of P r , ., , ^^.^
secant 6 = — — : =  (provided x^O),
abscissa of P x
. , /, abscissa of P a? , • i •, , r^.
cotangent 6 = — =  (provided y^O).
ordinate oi P y
These functions are written esc 6, sec 0, ctn d. From the
definitions follow directly the relations
csce = ^i, sece = , ctn0 =
sin 6' COS0' tan 9'
or
esc ^ • sin ^ = 1, sec ^ • cos ^ = 1, ctn • tan ^ = 1.
To the above functions may be added versed sine (written versin),
the coversed sine (written coversin), and the external secant (written
VI, § 116] TRIGONOMETRIC FUNCTIONS
169
exsec), which are defined by the equations versin 6 = 1 — coa 6, coversin e
= 1 — sin 6, and exsec 6 = sec d — 1.
It is left as an exercise to trace the variation of esc By sec 6,
ctn 0, as varies from 0° to 360°. Be careful to note tliat
ctn 0°, ctn 180°, esc 0°, esc 180°, sec 90°, sec 270° are undefined.
Why?
116. The Representation of the Functions by Lines. We
have seen in §§ 109111, that if we take a unit circle we may
represent sin 9, cos 0, and tan by means of lines. We will
now extend this representation to include esc 6, sec 6, ctn 6.
Fig. 107
Figure 107 shows the functions in a unit circle for an angle
6 in the first quadrant. We have
MF = sin e
OM=Gos6
AT
BS
tan^
ctn^
Or=sec^
OS = CSC $
Draw similar figures for angles in each of the other quad
rants. The points may be so labeled that the results given
for the first quadrant hold in any quadrant.
170 MATHEMATICAL ANALYSIS [VI, § 117
117. Relations among the Trigonometric Functions. As
one might imagine, the six trigonometric functions sine, cosine,
tangent, cosecant, secant, cotangent are connected by certain
relations. We shall now find some of these relations.
From Fig. 80 (§ 102) it is seen that for all cases we have
(1) • 0:2 4 2/2 = ,.2_
If we divide both sides of (1) by r^, we have
^ f ^ = 1 (by hypothesis r^0)\
or
sin2 + cos'e = l.
Dividing both sides by a;^, we have
1 + ^ = ^ iiix^O),
x^ x^
Therefore
1 + tan^ 6 = sec2 6.
Similarly dividing both sides of (1) by if gives
^ + 1 = S (if.'/^O);
or
ctn2 e + 1 = csc2 e.
Moreover, we have
X X COS
r
and, similarly,
ctne=5?i.
sm6
VI, § 118] TRIGONOMETRIC FUNCTIONS 171
118. Identities. By means of the relations just proved
any expression containing trigonometric functions may be
put into a number of different forms. It is often of the
greatest importance to notice that two expressions, although
of a different form, are nevertheless identical in value. (See
§ 47 for the definition of an identity.)
The truth of an identity is usually established by reducing
both sides, either to the same expression, or to two expres
sions which we know to be identical. The following examples
will illustrate the methods used.
Example 1. Prove the relation sec^ d + csc^ d = sec^ d csc^ d.
We may write the given equation in the form
sec2 d csc2 5,
or
which reduces to
cos'^ d sin^
?Hl!i±^2sif = sec2ecsc2^,
cos2 dsin^d
sec2 d csc2 d,
cos2 e sin2 d
sec2 d csc2 d = sec2 d csc^ 6.
Since this is an identity, it follows, by retracing the steps, that the
given equality is identically true.
Both members of the given equality are undefined for the angles 0°, 90°,
180°, 270°, 360° or any multiples of these angles.
cos*^ 6
Example 2. Prove the identity 1 + sin ^ = ■, — •
1 — sm ^
Since cos^ ^ = 1 — sin2 ^, we may write the given equation in the form
1 + sin = ^ ~ ^^"^ ^ or 1 + sin d = 1 + sin 6.
1  sin ^
As in Example 1, this shows that the given equality is identically true.
The righthand member has no meaning when sin = 1, while the left
hand member is defined for all angles. We have, therefore, proved that
the two members are equal except for the angle 90° or (4 ji  1) 90°, where
n is any integer.
172 MATHEMATICAL ANALYSIS [VI, § 118
The formulas of § 117 may be used to solve examples of the
type given in § 107.
Example 3. Given that sin d = /^ and that tan 6 is negative, find the
values of the other trigonometric functions.
Since sin^ d + cos^ ^ = 1, it follows that cos ^ = ± ^, but since tan 6 is
negative, 6 lies in the second quadrant and cos 6 must be — . More
over, tlie relation tan d = sin 0/ cos 6 gives tan 6 =— j%. The reciprocals
of these functions give sec ^ = — f, esc 6 = ^^^ ctn ^ = — ^.
EXERCISES
1. Define secant of an angle ; cosecant; cotangent.
2. Are there any angles for which the secant is undefined ? If so,
what are the angles ? Answer the same questions for cosecant and co
tangent.
3. Define versed sine ; coversed sine.
4. Complete the following formulas :
sin20 + cos2 0=? l + tan2^ = ? l+ctn2^=? tan5=?
Do these formulas hold for all angles ?
5. In what quadrants is the secant positive ? negative ? the cosecant
positive ? negative ? cotangent positive ? negative ?
6. Is there an angle whose tangent is positive and whose cotangent is
negative ?
7. In what quadrant is an angle situated if we know that
(a) its sine is positive and its cotangent is negative ?
(6) its tangent is negative and its secant is positive ?
(c) its cotangent is positive and its cosecant is negative ?
8. Express sin^ ^ + cos ^ so that it shall contain no trigonometric
function except cos 6.
9. Transform (1 + ctn^ 6) esc so that it shall contain only sin 6.
10. Which of the trigonometric functions are never less than one in
absolute value ?
11. For what angles is the following equation true : tan = ctn ?
12. How many degrees are there in when ctn ^ = 1 ? ctn^ = — 1 ?
sec ^ = \/2 ? CSC = \/2 ?
13. Determine from a figure the values of the secant, cosecants **»^
cotangent of 30°, 160^, 2W\ 330°.
VI, § 118] TRIGONOMETRIC FUNCTIONS
173
14. Determine from a figure the values of the secant, cosecant, and
cotangent of 45% 135°, 226% 316°.
16. Determine from a figure the values of the sine, cosine, tangent,
secant, cosecant, and cotangent of 60^, 120°, 240°, 300°.
16. Show that the graphs of the function sec e, esc ^, ctn d have the
forms indicated in the adjacent figures.
Prove the following identities and state for each the exceptional values
of the variables, if any, for which one or both members are undefined :
17. cos d tan d = sin d.
18. sin 6 ctn Q = cos 6.
1 + sin g _ cos5
cos 6
19
1 — sin g
20. sin2 6 — cos2 6 = 2 sin2 51.
21. ( 1  sin2 5) csc2 6 = ctn2 ^.
22. tan 6 + ctn 5 = sec 5 esc 0.
23. [x sinet y cos ey h [xcosdy sin 6^ = x^ + ^.
CSC 6
24.
= cos 0.
tan + ctn
26. 1  ctn* = 2 csc2  esc* 0.
26. tan2 5sin2 5 = tan2 5sin2g.
27. 2(1 + sin 0) ( 1 + cos 0) = (1 + sin + cos ey.
28. sin» + cos« 5 = 1—3 sin2 cos2 6.
CSC 5
29. _^!5^ +
30.
CSC 5 — 1 CSC 5 + 1
1 — tan 9 _ Ctn g  1
1 + tan 5 ctD 5 f 1
= 2 sec2 e.
174 MATHEMATICAL ANALYSIS [VI, § 118
31. [1 + tan e + sec e][l + ctn e — esc ^] = 2.
32. (taiid + sec^)2 = LiLSEi.
^ ^ 1  sin ^
33. CSC* ^ (1  cos* d)2 ctn2 ^ = 1.
34. (tan d — ctn ^)sin ^ cos = 1  2 cos^ e.
36. sec g  tan j ^ i _ 2 sec g tan ^ + 2 tan^ d.
sec + tan 6
36. ^^?i«±J^^5j = tanatan/3.
ctn a + ctn /3
37. sin (sec d + esc ^) — cos (sec ^ — esc 6) = sec ^ esc d.
Find algebraically the other trigonometric functions of the angle
when
38. ctn = 4: and sin is negative.
39. sin = I and sec is positive.
40. sec ^ = 2 and tan is negative.
41. CSC ^ = — 5 and ctn is positive.
119. Trigonometric Equations. An identity, as we have
seen (§ 47), is an equality between tvro expressions which is
satisfied for all values of the variables for which both expres
sions are defined. If the equality is not satisfied for all
values of the variables for which each side is defined, it is
called a conditional equality, or simply an equation. Thus
1 — cos ^ = is true only if ^ = w • 360°, where n is an integer.
To solve a trigonometric equation, i.e. to find the values of
for which the equality is true, we usually proceed as follows.
1. Express all the trigonometric functions involved in terms
of one trigonometric function of the same angle.
2. Find the value (or values) of this function by ordinary
algebraic methods.
3. Find the angles between 0° and 360° which correspond to
the values found. These angles are called particular solutions.
4. Give the general solution by adding n • 360°, where n is
any integer, to the particular solutions.
VI, § 119] TRIGONOMETRIC FUNCTIONS 175
Example 1. Find d when sin 5 = ^.
The particular solutions are 30° and 150°. The general solutions are
30° + w . 360°, 150° + n • 360°.
Example 2. Solve the equation tan ^ sin ^ — sin ^ = 0.
Factoring the expression, we have sin 6 (tan^— 1) = 0. Hence we
have sin ^ = 0, or tan ^—1=0. Why ?
The particular solutions are therefore 0°, 180°, 45°, 225°. The general
solutions are n • 360°, 180° + w • 360°, 45° + n . 360°, 225° + n . 360°.
Example 3. Find d when tan 6 + ctn 6 = 2.
The given equation may be written
tan (? + —^ = 2,
tan d
or
tan2 ^2 tan ^+1 = 0;
therefore
(tan ^  1)2 = 0, or tan tf = 1.
It follows that 6 = 46° or 225° ; or, in general,
^ = 46° + n . 360° or 226° + n • 360°.
EXERCISES
Give the particular and the general solutions of the following equations :
1. sin d = ^. 9. tan ^ =  1.
2. sin 5 =  ^. 10. ctn ^ =  1.
3. cos e= ^. 11. tan 6=1.
4. cos 5 = — ^. 12. ctn ^ = 1.
6. sec 6 = 2. 13. tan2 6 = 3.
6. sec ^ = — 2. 14. sin ^ = 0.
7. CSC 6 = 2. 15. cos ^ = 0.
8. esc ^ =  2. 16. tan 6 = 0.
Solve the following equations giving the particular and the general
solutions in each case :
17. sin d = cos 6. Ans. 45°, 226° ; 45° + n • 360°, 226° + n • 360°.
18. tan^^H 2sec2^ = 6.
19. 6 sin ^ + 2 cos2 6 = 5. Ans. 90° ; 90° + n • 360°.
20. cos2 ^ + 5 sin = 7.
176
MATHEMATICAL ANALYSIS [VI, § 119
21. 4 sin 5 — 3 esc ^ = 0.
22. 2 sin 6 cos^ 5 = sin 5.
23. cos 5 + sec 5 = f .
24. 2 sin = tan d. Ans. Particular solutions : 0'', 180°, 60°, 300°.
25. 3 sin ^ + 2 cos ^ = 2.
26. 2cos2 ^1 = 1 sin2 e .
120. The Trigonometric Functions of — 9. Draw the angles
and — 9, where OP is the terminal line of and OP' is the
terminal line of — 0. Figure 108 shows an angle in each of
P P
X V
p7>
^ ^
Fig. 108
X
V X
the four quadrants. We shall choose OP — OP' and {x, y) as
the coordinates of P and (x\ y') as the coordinates of P'. In
all four figures
Hence
x' = ic, y' — — y, r' — T.
sin(^)=^ = :=^ = sin(9,
r r
cos ((9)=^ = = cos d,
r r
tan(^) = ^ = ^:^ = tand!
Also,
csc(— ^)= — csc^; sec (— ^)= seed; ctn ( d)=— ctnd.
VI, § 121] TRIGONOMETRIC FUNCTIONS
177
121. The Trigonometric Functions of 90° — 6. Figure 109
represents angles 9 and 90° — 0, when ^ is in each of the f oui
quadrants. Let OP be the terminal line of 6 and OP', the
Y
^
»'
90'9
VF
X' J
^ X
Fio. 109
terminal line of 90°  6. Take OP' = OP and let (x, y) be the
coordinates of P and (a;', y') the coordinate of P. Then in all
four j&gures we have
^ = y> y'^^y r' = r.
Hence
sin(9O°0) = 4= = cos^,
r r
tan (90°^)
— ^ — IL —
r r
= ^ = ?=ctnd.
cos(90°^) =  = ^=sind,
r r
y
Also,
CSC (90° ^)=sec^,
sec (90° (9)=csc^;
ctn(90° ^)=tand.
Definition. The sine and cosine, the tangent and cotan
gent, the secant and cosecant, are called cofunctions of each
other.
The above results may be stated as follows : Any function
of an angle is equal to the corresponding cofunction of the com
plementary angle.*
*Two angles are said to be complementary if their sum is 90°, regardless of
the size of the angles.
178 MATHEMATICAL ANALYSIS [VI, § 122
122. The Trigonometric Functions of 180° — 0. By draw
ing figures as in §§ 120, 121, the following relations may be
proved :
sin (180°  0) = sin 0, esc (180° 6)= esc 0,
cos (180° 0) = — cos e, sec (180° 6) =  sec 6,
tan (180°  (9) =  tan 0, ctn (180° &)= ctn 6.
The proof is left as an exercise.
123. The Trigonometric Functions of 180° + 6. Similarly,
the following relations hold :
sin (180° + ^) =  sin 0, esc (180° + 0) =  esc 6,
cos (180° 4 ^) =  cos 0, sec (180° + ^) =  sec 6,
tan (180° + (9) = tan B, ctn (180° + ^) = ctn d.
The proof is left as an exercise.
124. Summary. An inspection of the results of §§ 120123
shows :
1. Each function of — B or 180° ± B is equal in absolute value
(but not always in sign) to the same function of 0,
2. Each function of 90° — B is equal in magnitude and in sign
to the corresponding cofunction of 6.
These principles enable us to find the value of any function
of any angle in terms of a function of a positive acute angle
(not greater than 45° if desired) as the following examples
show.
Example 1. Reduce cos 200° to a function of an angle less than 45°.
Since 200° is in the second quadrant, cos 200° is negative. Hence
cos 200°=  cos 20°. Why ?
Example 2. Reduce tan 260° to a function of an angle less than 45°.
Since 260° is in the third quadrant, tan 260° is positive. Hence
tan 260° = tan 80° = ctn 10° (§ 121).
VI, § 124] TRIGONOMETRIC FUNCTIONS 179
EXERCISES
Reduce to a function of an angle not greater than 45° :
1. sin 163°. 6. esc 900°.
2. cos( 110°). 6. ctn ( 1215°).
Ans. — cos 70° or — sin 20°. 7, tan 840°.
3. sec (265°). 8. sin 510°.
4. tan 428°.
Find without the use of tables the values of the following functions :
9. cos 570°.
13. cos 150°.
10. sin 330°.
11. tan 390°. W tan 300°.
12. sin 420°.
Reduce the following to functions of positive acute angles :
16. sin 250°. 18. sec (245°).
Ans. — sin 70° or — cos 20°. 19, esc (— 321°).
16. cos 158°. 20. sin 269°.
17. tan (389°).
21. Prove the following relations from a figure :
(a) sin (90° + d) = cos d. (c) sin (180° + 6)=  sin $.
cos (90° + e) =  sin d. cos (180° + ^) =  cos 0.
tan (90° + d) =  ctn 0. tan (180° + 0) = tan 0.
esc (90° + 0) = sec 0. CSC (180° + ^) =  esc ^.
sec (90° + ^) =  CSC 0. sec (180° + 0) = 8ec0.
Ctn (90° + 0)= tan 0. ctn (180° \0) = ctn 0.
(6) sin (180° 0)= sin 0. (d) sin (270°  ^) =  cos 0.
cos (180° 0) =  cos 0. cos (270° 0) =  sin 0.
tan ( 1 80°  ^) =  tan 0. tan (270°  ^) = ctn 0.
CSC (180°  ^) = CSC 0. CSC (270°  ^) =  sec 0,
sec (180° 0) =  sec 0. sec (270° 0) =  esc 0.
ctn (180°  ^) =  ctn ^. ctn (270°  ^) = tan 0.
(e) sin (270° + ^) =  cos ^.
cos (270° + ^)= sin ^.
tan (270° + 0) = — ctn 0.
CSC (270° + ^) =  sec ^.
sec (270° + 0)= CSC ^.
ctn(270° + ^) = ~ tan^.
180
MATHEMATICAL ANALYSIS
[VI, § 125
125. Law of Sines. Consider any triangle ABC with the
altitude CD drawn from the vertex C (Fig. 110).
I) B D A
Fia. 110
In all cases we have sin A= , sin B =
b
a
Therefore, dividing, we obtain
sin^ a
or
sin^ 6'
a b
(2)
sin A sin B
If the perpendicular were dropped from B, the same argu
ment would give
^ = ^. (3)
sin A sin C
Combining results (2) and (3) we have
a _ b _ c
sin A sin 5 sin C*
This law is known as the law of sines and may be stated as
follows :
Amj tivo sides of a triangle are proportional to the sines of the
angles opposite these side'i.
126. Law of Cosines. Consider any triangle ABC with the
altitude CD drawn from the vertex C (Fig. 111).
In Fig. Ill a
AD = 6 cos ^ ; CD = 6 sin ^ ; DB = c — beosA.
In Fig. Ill b
AD = — 6 cos 4 ; CD = 6 sin ^ ; DB = c — b cos A
VI, § 127] TRIGONOMETRIC FUNCTIONS
In both figures
Therefore
a2 = c2 — 2 6c cos A{¥ cos^ A\ll^ sin^ A
= c^ — 2 &c cos ^ 4 (cos2 A + sin* A) b%
181
b
D B D
whence
FlQ. Ill
a2 = &2 + c2 — 2 6c cos A,
Similarly it may be shown that
52 = c2 f a2  2 ca • cos B,
c^ z= 0} ■\ h"^ — 2 ah • cos C.
Any one of these similar results is called the law of cosines
It may be stated as follows :
Tlie square of any side of a triangle is equal to the sum of the
squares of the other two sides diminished by tivice the product of
these two sides times the cosine of their included angle.*
127. Solution of Triangles. To solve a triangle is to find
the parts not given, when certain parts are given. From
geometry we know that a triangle is in general determined
when three parts of the triangle, one of which is a side,
* Of what three theorems in elementary geometry is this the equivalent ?
182
MATHEMATICAL ANALYSIS
[VI, § 127
are given.* Eight triangles have already been solved
(§ 106 £f.), and we shall now make use of the laws of sines and
cosines to solve oblique triangles. TJie methods employed
will be illustrated by some examples. It will be found
advantageous to construct the triangle to scale, for by so doing
one can often detect errors which may have been made.
128. Illustrative Examples.
Example 1. Solve the triangle ABC, given
^^^^^ A = 30^^ 20', B = 60° 45', a = 276.
^A Solution :
C = 180°  (^ + B)=1S0°  91° 5' = 88° 55' ;
a sin B 276 sin 60=
sin J.
sin 30= 20
45^ ^ (276) (0.8725) ^^ygQ.
I' 0.5050 ' '
also
a sin G ^ 276 sin 38° 55' ^ (276) (0.i)998) ^ ^^g ^
Check : It is left as an exercise to show that for these values we have
c2 = a2 + &2 _ 2 a6 cos C.
Example 2. Solve the triangle ABC, given
A = 30°, 6 = 10, a = 6. ^Q^
Constructing the triangle ABC, we see that .^x^so"
two triangles AB\ C and AB2C answer the descrip ^
tion since b>a> altitude CD.
Solution : Now
Fig. 113
whence
But
B2
?H^ = ^orsin5i=^^Hl^
sin ^ a a
Bi = 56.5°.
180°  Bi = 180°  56.5° = 123.6°,
= 0.833,
and
Ci = 180°  (^ + Bi) = 180°  86.5° = 93.6°,
Ci = 180° (A + Bi) = 180°  153.5° = 26.6°.
* When two sides and an angle opposite one of them are given, the triangle
is not always determined. Why ?
VI, § 128] TRIGONOMETRIC FUNCTIONS 183
Now
Also
cg^ sin (72 ^ oj. c^^ a sin C^ ^ (6) (0.446) ^ g g^^
a sin ^ ' sin ^ 0.600
Ci^ sinCi Qj, asm(7i^ (6)(0.998) ^^^Q3
a sin^' sin J. 0.600
Check : Cy^ = a^ \ b^ — 2 ah cos (7i.
143.5 = 86+ 100 +(2)(6)(10)(0.061) = 143.3.
C22 = a2 4 62 _ 2 a6 cos O2.
28.62 = 36 + 100(2)(6)(10)(0;896)  28.60.
Example 3. Solve the triangle ABC, given a= 10, 6=6, 0=40°.
Solution : c^ = a^ + 6^ _ 2 a6 cos C j^
= 100 + 36  ( 120) (0. 766) = 44.08.
Therefore c = 6.64. Now \fy^ \V
sin .4 = «^^5iZ = ^lOHO643) ^ ^ ggg
c 6.64 '
i.e. 4 = 104.6°. Likewise
sin5 = ^«l5_^=(^KM43)^0.581,
c 6.64
Check : A + B^ G= 180.0°.
Example 4. Solve the triangle ABC when
a = 7, 6 = 3, c = 6.
From the law of coshies,
= != 0.500,
2 6c 2
^ + c262 ^13^0Q^8,
2ac 14
^2 + 62  C2 11 n 7QA
COS C = = — = 0. 786.
2 a6 14
Therefore ^ ^ 120°, 5 = 21.8°, C = 38.2°.
Check . A^B+ C= 160.0°.
EXERCISES
Solve the triangle ABC, given
(a) ^ = 30°, 5 = 70°, a = 100;
(6)^ = 40°, 5 = 70°, c = 110;
(c) A = 45.5°, = 68.6°, 6 = 40 ;
(d)J5 = 60.5°, C = 44°20', c = 20 ;
184 MATHEMATICAL ANALYSIS [VI. § 128
(e) a = 30, 6 = 64, 0=50°; (g)a=10, 6 = 12, c = U;
(/) 6 = 8, a = 10, O = 60° ; {h) a = 21, 6 = 24, c = 28.
2. Determine the number of solutions of the triangle ABO when
(a) A = 30°, 6 = 100, a = 70
(e)
A =
30°,
6 =
100,
a =
120
(/)
A =
106°,
6 =
120,
a =
16
ig)
A =
90°,
6 =
15,
a =
14.
(6) A = 30°, 6 = 100, a = 100
(c) ^ = 30°, 6 = 100, a = 50
(d) A = 30°, 6 = 100, a = 40
3. Solve the triangle ABC when
(a) ^ = 37° 20', a = 20, 6 = 26 ; (c) ^ = 30°, a = 22, 6 = 34.
(6) ^ = 37° 20', a = 40, 6 = 26;
4. In order to find the distance from a point A to z. point JB, a line
AC and the angles CAB and ACB were measured and found to be
300 yd., 60° 30', 56° 10' respectively. Find the distance AB.
5. In a parallelogram one side is 40 and one diagonal 90. The angle
between the diagonals (opposite the side 40) is 25°. Find the length of
the other diagonal and the other side. How many solutions ?
6. Two observers 4 miles apart, facing each other, find that the angles
of elevation of a balloon in the same vertical plane with themselves are
60° and 40° respectively. Find the distance from the balloon to each
observer and the height of the balloon.
7. Two stakes A and B are on opposite sides of a stream ; a third
stake C is set 100 feet from A, and the angles ACB and CAB are observed
to be 40° and 110°, respectively. How far is it from Ato B?
8. The angle between the directions of two forces is 60°. One force
is 10 pounds and the resultant of the two forces is 15 pounds. Find the
other force.*
9. Resolve a force of 90 pounds into two equal components whose
directions make an angle of 60° with each other.
10. An object B is wholly inaccessible and invisible from a certain
point A. However, two points C and D on a line with A may be found
such that from these points B is visible. If it is found that CD =
800 feet, CA = 120 feet, angle DCB = 70°, angle CDB = 50°, find the
length AB.
* It is shown in physics that If the line segments AB
3 ~ ^ ^ ^^^ ^^ represent in magnitude and direction two forces
r ^^^/ acting at a point A, then the diagonal AJ) of the parallelo
fy^^ / gram ABCD represents both in magnitude and direction
A B the resultant of the two given forces.
VI, § 128] TRIGONOMETRIC FUNCTIONS
185
11. Given a, 6, A, in the triangle ABC. Show that the number of
possible solutions are as follows :
[ a < 6 sin ^ no solution,
I bsmA<ia<.b two solutions,
\a>b
I a = 6 sin
J
one solution.
( a <b no solution,
I o > 6 one solution.
12. The diagonals of a parallelogram are 14 and 16 and form an angle
of 50°. Find the length of the sides.
13. Resolve a force of magnitude 150 into two components of 100 and
80 and find the angle between these components.
14. It is sometimes desirable in surveying to extend a line such as AB
in the adjoining figure. Show that this can be done by means of the
broken line ABODE. What measurements are necessary ?
15. Three circles of radii 2, 6, 5 are mutually tangent. Find the angles
between their lines of centers.
16. In order to find the distance between two objects A and B on op
posite sides of a house, a station C was chosen, and the distances CA
= 500 ft., CB = 200 ft., together with the angle ACB  65° SO' were
measured. Find the distance from A to B.
17. The sides of a field are 10, 8, and 12
rods respectively. Find the angle opposite the
longer side.
18. From a tower 80 feet high, two objects,
A and B, in the plane of the base are found to
have angles of depression of 13° and 10° respec
tively ; the horiz(mtal angle subtended by A and B at the foot C of the
tower is 44°. Find the distance from A to B.
186 MATHEMATICAL ANALYSIS [VI, § 129
129. Areas of Oblique Triangles.
1. When two sides and the included angle are given.
Denoting the area by S^ we have from geometry
G S = ^ch,
but ^ = & sin ^ ; therefore
(4) S^^cb sin A.
Likewise,
Fig. 116
S — ^ab sin O and S = ^acsinB.
2. When a side and two adjacent angles are given.
Suppose the side a and the adjacent angles B and C to be
given. We have just seen that S = ^ac sin B. But from the
law of sines we have
a sin C
Therefore
S =
SIR A
a^ • sin jB • sin C
2 sin A
But sin A = sin [180°  (5 + (7)] = sin (5 + C). Therefore
^ _ a'^ sin B sin O
~ 2 sin (5+0)'
3. When the three sides are given.
We have seen that S = ^bc sin A. Squaring both sides of
this formula and transforming, we have
™ 7)2,.2 7)2^2
S^ = ^ sin2^ = ^(lcos2^)
whence,
= (l4cos^).(lcos^);
^^A 1 b^\c^a'
S' = ^ 1h
2 V 2 6c
;N bcf. b^ + c^a^ \
J 2[ 2 be J
VI, § 130] TRIGONOMETRIC FUNCTIONS 187
4 ' 4
_M_c4;_a b \c—a a~b\c gf 6~c
2*2*2*2'
which may be written in the form
S^ = s(sa){sb)(sc),
where 2s = a + &4c. Therefore,
(5) S = Vs(sa)(s6)(sc).
130. The Radius of the Inscribed Circle. If r is the radius
of the inscribed circle, we have from elementary geometry,
since s is half the perimeter of the triangle, S = rs ; equating
this value of S to that found in equation (5) of the last article
and then solving for r, we get,
^J (sa)(sb)(sc) _
^ 5
EXERCISES
Find the area of the triangle ABC, given
1. a = 26, b = 31.4, C = 80° 25'. 4. a = 10, 6 = 7, = 60°.
2. 6 = 24, c = 34 3, ^ = 60° 25'. 5. a = 10, 6 = 12, C=60°.
3. a = 37, 6 = 13, C = 40°. 6. a = 10, 6 = 12, C = 8°.
7. Find the area of a parallelogram in terms of two adjacent sides
and the included angle.
8. The base of an isosceles triangle is 20 ft. and the area is 100/ \/3
sq. ft. Find the angles of the triangle. Ans. 30°, 30°, 120°.
9. Find the radius of the inscribed circle of the triangle whose sides
are 12, 10, 8.
10. How many acres are there in a triangular field having one of its
sides 60 rods in length and the two adjacent angles, respectively, 70°
and 60° ?
CHAPTER VII
TRIGONOMETRIC RELATIONS
131. Radian Measure. In certain kinds of work it is more
convenient in measuring angles to use, instead of the degree,
a unit called the radian. A radian is defined as the angle at
the center of a circle whose subtended arc is equal in length
to the radius of the circle (Fig. 117). Therefore, if an angle 6
at the center of a circle of radius r units subtends an arc of
s units, the measure of in radians is
(1) 0=i.
Since the length of the whole circle is 2 irr, it follows that
^^ = 2 TT radians = 360°,
r
(2) IT radians = 180°.
Fig. 117 Therefore,
IQAO
1 radian = ±^ = 57° 17' 45" (approximately).
TT
It is important to note that the radian * as defined is a con
stant angle, i.e., it is the same for all circles, and can therefore
be used as a unit of measure.
* The symbol »" is often used to denote radians. Thus 2" stands for 2
radians, tt* for tt radians, etc. When the angle is expressed in terms of t (the
radian being the unit) , it is customary to omit »•. Thus, when we refer to an
angle tt, we mean an angle of tt radians. When the word radian is omitted, it
should be mentally supplied in order to avoid the error of supposing ir means
180. Here, as in geometry, IT = ;i. 14159. ...
188
VII, § 132] TRIGONOMETRIC RELATIONS 189
Erom relation (2) it follows that to convert radians into
degrees it is only necessary to multiply the number of radians
by ISO/V, while to convert degrees into radians we multiply
the number of degrees by 7r/180. Thus 45° is 7r/4 radians ;
7r/2 radians is 90°.
132. The Length of Arc of a Circle. From relation (1),
§ 131, it follows that s=re
5 = re.
That is (Fig. 118), if a central angle is measured
in radians, and if its intercepted arc and the
radius of the circle are measured in terms of
the same unit, then ^°'
length of arc = radius x central angle in radians.
EXERCISES
1. Express the following angles in radians :
25°, 145°, 225°, 300°, 270°, 450°, 1150°.
2. Express in degrees the following angles :
IT 7 IT Sir n 57r
4' "T' T'^'^'T*
t. A circle has a radius of 20 inches. How many radians are there in
an angle at the center subtended by an arc of 25 inches ? How many
degrees are there in this same angle ? Ans. f ; 71° 37' approx.
4. Find the radius of a circle in which an arc 12 inches long subtends
an angle of 35°.
6. The minute hand of a clock is 4 feet long. How far does its ex
tremity move in 22 minutes ? 
6. In how many hours is a point on the equator carried by the rotation
of the earth on its axis through a distance equal to the diameter of the earth ?
7. A train is traveling at the rate of 10 miles per hour .on a curve of
half a mile radius. Through what angle has it turned in one minute ?
8. A wheel 10 inches in diameter is belted to a wheel 3 inches in
diameter. If the first wheel rotates at the rate of 5 revolutions per
minute, at what rate is the second rotating ? How fast must the former
rotate in order to produce 6000 revolutions per minute in the latter ?
190 MATHEMATICAL ANALYSIS [VII, § 133
133. Angular Measurement in Artillery Service. The
divided circles by means of which the guns of the United States Field
Artillery are aimed are graduated neither in degrees nor in radians, but
in units called mils. The mil is defined as an angle subtended by an arc
of ■g:^^ of the circumference, and is therefore equal to
2_ir_ ^ SAAW ^ 0.00098175 = (0.001  0.00001825) radian.
6400 3200 ^
The mil is therefore approximately one thousandth of a radian.
(Hence its name.)*
Since (§132)
length of arc = radius x central angle in radians,
it follows that we have approximately
T5) li ill A
length of arc = x central angle in mils ;
1000 ^ '
i.e. length of arc in yards = (radius in thousands of yards) • (angle
in mils). The error here is about 2 %.
Example 1. A battery occupies a front of 60 yd. If it is at
5500 yd. range, what angle does it subtend (Fig. 119)? We
have, evidently,
angle = — = 11 mils.
5.5
Example 2. Indirect Fire, t A battery posted with its right gun at G
is to open fire on a battery at a point T, distant 2000 yd. and invisible
* To give an idea of the value in mils of certain angles the following has
been taken from the Drill Regulations for Field Artillery (1911), p. 164:
" Hold the hand vertically, palm outward, arm fully extended to the front.
Then the angle subtended by the
width of thumb is 40 mils
width of first finger at second joint is 40 mils
width of second finger at second joint is .... 40 mils
width of third finger at second joint is 35 mils
width of little finger at second joint is 30 mils
width of first, second, and third fingers at second joint is . 115 mils
These are average values."
t The limits of this text preclude giving more than a single illustration of
the problems arising in artillery practice. For other problems the student is
referred to the Drill Regulations for Field Artillery (1911), pp. 57, 61, 150164;
and to Andrews, Fundamentals of Military Service, pp. 153159, from which
latter text the above example is taken.
VII, § 133] TRIGONOMETRIC RELATIONS
191
from Cr (Fig. 120) . The officer directing the fire takes post at a point
B from which both the target T ami a church spire P, distant 3000 yd.
from O are visible. B is 100 yd. at the right of the line &T and 120 yd.
at the right of the line GP and the officer finds by measurement that the
angle PBT contains 3145 mils. In order to train the gun on the target
the gunner must set off the angle PGT on jf rpr
the sight of the piece and then move the gun
until the spire P is visible through the sight.
When this is effected, the gun is aimed at T.
Let F and E be the feet of the perpen
diculars from B to G^Tand G^P respectively,
and let BV and BP' be the parallels to
OT and OP that pass through B. Then,
evidently, if the officer at B measures the
angle PPT, which would be used instead
of angle POT were the gun at B instead
of at G^ and determines the angles TBT' =
FTB and PBP' = EPB, he can find the
angle PGT from the relation
PGT=P'BT'= PBT
Fig. 120
TBT' PBP'.
Now
tan FTB
FB
TF
tan EPB = — .
PF
Furthermore if FTB and EPB are small angles, i.e., if FB and EB are
small compared with OT Sind OP respectively, the radian measure of the
angle is approximately equal to the tangent of the angle. Why ? Hence
we have
FB
GT
EB
OP
100
Therefore
FTB  tan FTB =
EPB = tan EPB
approximately.
TBT' = FTB = ^^^ radians
2000
50 mils.
PBP'
EPB = ^ radians
3000
40 mils.
Hence POT = PBT  TBT  PBP*
= 3145  60  40
— 3056 mils,
which is the angle to be set off on the sight of the gun.
1^2 MATHEMATICAL ANALYSIS [VII, § 133
Hence for the situation indicated in Fig. 120 we have the following
rule : *
(]) Measure in mils the angle PBT from the aiming point P to the
target T as seen at B.
■ : (2) Measure or estimate the offsets FB and EB in yards, the range
G^Tand the distance GB of the aiming point P in thousands of yards.
(3) Compute in mils the offset angles by means of the relations
TBT = FTB,
FBP' = EPB,
TBr=—,
GT'
CrF
(4) Then the angle of deflection FGT is equal to the angle FBT
diminished by the sum of the offset angles.
EXERCISES
1. A battery occupies a front of 80 yd. It is at 5000 yd. range.
What angle does it subtend ?
2. In Fig. 120 suppose FBT= 3000 mils, FB = 200 yd., GT = 3000 yd.,
EB = 150 yd., GP = 4000 yd. Find the number of mils in FGT.
3. A battery at a point G is ordered to take a masked position and be
ready to fire on an indicated hostile battery at a point T whose range is
known to be 2100 yd. The battery commander finds an observing station
B, 200 yd. at the right and on the prolongation of the battery front, and
175 yd. at the right of PGr An aiming point P, 5900 yd. in the rear, is
found, and PBT is found to be 2600 mils. Find FGT.
134. Inverse Trigonometric Functions. The equation
X = sin y (1)
may be read :
y is an angle whose sine is equal to a;,
a statement which is usually written in the contracted, form
y — arc sinflj.f (2)
* There are three cases with corresponding rules, depending on whether P
is in front of, rear of, or on the flank of G.
t Sometimes written y = sini.c. Hefe — 1 is not an algebraic exponent,
but merely a part of a functional symbol. When we wish to raise sin x to
the power — 1, we write (sin a;)i.
VII, § 134] TRIGONOMETRIC RELATIONS
193
For example, x = sin 30° means that x = ^, while y = arc sin^
means that y = 30°, 150°, or in general (n being an integer),
30° + n . 360° ; 150° + n • 360°.
Since the sine is never greater than 1 and never less than
— 1, it follows that — 1 ^ a; ^ 1. It is evident that there is
an unlimited number of values of y = arc sin x for a given value
of X in this interval.
We shall now define the principal value Arc sin x* of arc sin x^
distinguished from arc sin x by the use of the capital A, to
be the numerically smallest angle whose sine is equal to x.
This function like arc sin x is defined only for those values of
X for whichsk
 1< a; < 1.
The difference between arc sin x and Arc sin x is well illus
trated by means of their graph. It is
evident that the graph oi y = arc sin x,
i.e. X = sin y is simply the sine curve
with the role of the x and y axes inter
changed. (See Fig. 121.) Then for every
admissible value of x, there is an un
limited number of values of y ; namely,
the ordinates of all the points Pi, P2, — , in
which a line at a distance x and parallel
to the 2/axis intersects the curve. The
singlevalued function Arc sin x is repre
sented by the part of the graph between
Jlf and iV.
Similarly arc cos x, defined as " an angle whose cosine is a;,"
* Sometimes written Sinicc, distinguished from sin^a; by the ase of the
capital S.
T
.
2t
^
Ps
( sir
ir
\
n
IT
i
>
y
N
i
^
1 X
M
G
y= arc sin x
y=
'Ar
(;sinx
Fig. 121
194
MATHEMATICAL ANALYSIS [VII, § 134
has an ■anlimited. number of values for
every admissible value of a; (— 1^ a; ^ 1).
We shall define the principal value Arc
cos X as the smallest positive angle whose
cosine is x. That is,
< Arc cos a? ^ TT.
Figure 122 represents the graph of
y = arc cos x and the portion of this graph
between M and N represents Arc cos x.
Similarly we write x = tan y as y,= arc
tan X, and in the same way we define the
symbols arc ctn x ; arc sec x ; arc esc x.
The principal values of all the inverse trigonometric functions
are given in the following table.
7
2ir
■)
3jr
Ps
N
^■J
IT
^
\
M
1
IT
<
1 X
y= arc cos x
y=Arc cosz
Fig. 122
y =
Arc sin x
Arc cos X
Arc tan x
Range of x
Range of y
X positive
X negative
^to^
2 2
1st Quad.
4th Quad.
l^x^l
to T
1st Quad.
2d Quad.
all real values
1st Quad.
4th Quad.
Arc ctn x
Arc sec x
Arc CSC X
Range of x
Range of y
X positive
X negative
all values
tOTT
1st Quad.
2d Quad.
a; > 1 or a; <  1
tOTT
1st Quad.
2d Quad.
X>l0TX<l
^ to"^
2 2
1st Quad.
4th Quad.
In so far as is possible we select the principal value of each
inverse function, and its range, so that the function is single
valued, continuous, and takes on all possible values. This ob
viously cannot be done for the Arc sec x and for Arc esc y.
VII, § 134] TRIGONOMETRIC RELATIONS 195
EXERCISES
1. Explain the difference between arc sin x and Arc sin x.
2. Find the values of the following expressions :
(a) Arc sin \. (&) arc sin \. (c) arc tan 1.
(d) Arc tan  1. (e) arc cos I^. (H Arc cos lA.
2 ' 2
3. What is meant by the angle tt ? 7r/4?
4. Through how many radians does the minute hand of a watch turn
in 30 minutes ? in one hour ? in one and one half hours ?
5. For what values of x are the following functions defined :
(a) arc sin x ? (6) arc cos cc? (c) arc tan x ?
id) arc ctn X ? (e) arc sec a; ? (/) arc esc cc ?
6. What is the range of values of the functions :
(a) Arc sin x ? (&) Arc cos x ? (c) Arc tan x ?
id) Arc ctn x ? (e) Arc sec x ? (/) Arc esc x ?
7. Draw the graph of the functions :
(a) arc sin x. (&) arc cos x. (c) arc tan x.
(<?) arc ctn X. (e) arc sec x. (/) arc esc x.
8. Find the value of cos (Arc tan ).
Hint. Let Arc tan \ = 6. Then tan ^ = f and we wish to find the
value of cos e.
9. Find the values of cos (arc tan ).
10. Find the value of the following expressions :
(a) sin (arc cos ). (c) cos (Arc cos y\). (e) sin (Arc sin \).
(&) sin (arc sec 3). {d) sec (Arc esc 2). (/) tan (Arc tan 6).
11. Prove that Arc sin (2/6) = Arc tan (2/V2T).
12. Find x when Arc cos (2 x^ — 2 x) = 2 7r/3.
Find the values of the following expressions :
13. cos [90^ — Arc tan ].
14. sec [90° Arc sec 2].
15. tan [90°  Arc sin ^{\.
196
MATHEMATICAL ANALYSIS [VII, § 135
135. Projection. Consider two directed lines p and g in a
plane, i.e. two lines on each of which one of the directions
has been specified as positive (Fig. 123). Let A and B be
any two points on p and let A\ B' be the points in which per
Fig. 123
pendiculars to q through A and B, respectively, meet q. The
directed segment A'B' is called the projection of the directed seg
ment AB on q and is denoted by
A'B' = proj^ AB.
In both figures AB is positive. In the first figure A'B' is posi
tive, while in the second figure it is negative.
As special cases of this definition we note the following :
1. If p and q are parallel and are directed in the same way,
we have
^m]^AB=:AB.
2. If p and q are parallel and are directed oppositely, we
have
^xo]^AB=AB.
3. If p is perpendicular to q, we have
proj,^B = 0.
It should be noted carefully that these propositions are true
no matter how A and B are situated on p.
We may now prove the following important proposition;
VII, § 135] TRIGONOMETRIC RELATIONS
197
If Aj B are any two points on a directed line p, and q is
any directed line in the same plane with p, then we have both in
Tnagnitvde and sign :
(1) proj\ AB = AB' cos {qp),
where (qp) represents an angle through ichich q must be rotated
in order to make its direction coincide with thedirection of p.
We note first that all possible determinations of the angle
(qp) have the same cosine, since any two of these determina
tions differ by multiples of 360° (Fig. 124). We shall prove
Fig. 124
the proposition first for the case where AB has the same direc
tion as p, i.e. where AB is positive. To this end we draw
through A (Fig. 125) a line qi parallel to q and directed in the
A'
\Z.
B'
^
A Bi
^
B
J.
I' J
S' '
^
Fig. 125
same way. (We may evidently assume without loss of gener
ality that q is horizontal and is directed to the right.)
Let A'B' have the same significance as before and let BB'
meet ^'i in Bi. Then, by the definition of the cosine, we have
A 7?
—^ = cos (q^p) = cos (qp)y
in magnitude and in sign ; or
ABi = ^5 cos (qp).
But
Therefore
AB, = A'B'=VTOj^AB.
projg AB = AB cos (qp).
198 MATHEMATICAL ANALYSIS [VII, § 135
Finally, if AB is negative, BA is positive, and, by the result
just obtained, we should have
B'A' == BA cos (qp).
Hence, changing signs on both sides of this equation, we
have
A'B' = AB cos (q 2^).
The special cases 1, 2, 3 listed on p. 196 are obtained from
formula (1) by placing (qp) equal to 0°, 180°, 90°, respectively ;
for cos 0° = 1, cos 180°   1, cos 90° = 0.
136. Application of Projection. In Physics, forces and
velocities are usually represented by line segments. A force
of 20 pounds, for example, is represented by a segment 20 units
in length and drawn in the direction of the force. A velocity
of 20 feet per second is represented by a segment 20 units in
length and drawn in the direction of the motion.
The projection on a given line I of a segment representing
a force or velocity represents the component of the force or
velocity in the direction of l.
Example. A smooth block is sliding down a smooth incline
which makes an angle of 30° with the horizontal. If the block
weighs 10 lb., what force acting directly up
the plane will keep the block at rest ?
Draw the segment AB 10 units in length,
directly downward to represent the force
exerted by the weight. Project this segment
Fig. 126 ^^ ^^^ incline and call this projection AC.
Now angle ABC = 30°. Therefore AC = AB sin 30° = 5. This
is the component of the force AB down the plane. Therefore
a force of 5 lb. acting up the plane will keep the body at rest.
A
•^
B
a! c"
B'
VII, § 136] TRIGONOMETRIC RELATIONS 199
Theorem. If A, B, C are any three points in a plarie, arid I
is any directed line iii the plane, the algebraic sum of the projec
tions of the segments AB and BC on I is equal to the projection of
the segment AC on I.
As a point traces out the path, from A to B, and then from
B to O (Fig. 127), the projection of the poiiit traces out the
segments from A' to B' and then from B'
to C. The net result of this motion is a
motion from A' to C which represents
the projection of AC, i.e.
A'B' + B'C = A'C, Fig. 127
EXERCISES
1. What is the projection of a line segment upon a line I, if the line
segment is perpendicular to the line I ?
2. Find proj^^ AB and proj^ AB* in each of the following cases, if a
denotes the angle from the ccaxis to AB.
(a) AB = 5, a = 60°. (c) AB = 6, a = 90°.
(6) AB = 10, a = 300°. (d) AB = 20, a = 210°.
3. Prove hy means of projection that in a triangle ABC
a = b cos C + c cos B.
4. If proja; AB = 3 and projy AB = — 4, find the length of AB.
6. A steamer is going northeast 20 miles per hour. How fast is it
going north ? going east ?
6. A 20 lb. block is sliding down a 15° incline. Find what force
acting directly up the plane will just hold the block, allowing one half a
pound for friction.
7. Prove that if the sides of a polygon are projected in order upon any
given line, the sum of these projections is zero.
* Proji AB and projy AB mean the projections of AB on the xaxis and
the yaxis, respectively.
200
MATHEMATICAL ANALYSIS [VII, § 137
137. Rotation in a Plane. Suppose that a point P{Xj y) in
a plane moves on the arc of a circle with center at the origin 0,
through an angle a. Suppose that its position after this
rotation is Pix^ y') referred to the same axes of coordinates.
We desire to find x' and y' in terms of a;, y, and a.
In Fig. 128 we have
drawn P and its coordi
nates X = OM, y = MP, and
the new position OM' P' of
the triangle OMP after a
rotation about the origin
through an angle a. The
coordinates x' = ON, ?/' =
NP' of P are the pro
jections of OP on the
a;axis and the yaxis re
spectively, and these pro
jections are equal respec
tively to the sum of the projections of OM' and M'P on the
respective axes. Hence,
x' = proj, OP = proj, OM' f proj, M'P
= OM' cos {OX, OM') 4 M'P cos (OX, M'P)
= X cos a \ y cos (a 1 7r/2)
= X cos a — y sin a.
y' = proj, OP = proj^ OM' f proj^ M'P'
= OM' cos (OF, 03/') f 3/'P' cos ( OY, M'P')
= X cos (— 7r/2 + «) f 2/ cos a
= X sin a h y cos a.
Therefore, if the point P{x, y) is rotated about the origin
through an angle a, the coordinates (x', y') of its new position
are given by the formulas
Y>
\
1
P'l
1 ^
N. CC
m
k
L /
P
hs\^
V
'/>'''x' \
1^
T X
¥ ^X
Fig. 128
VII, § 138] TRIGONOMETRIC RELATIONS
201
(1)
J x' = X cos a — y sin a
[j/ = x sin a + 1/ cos a.
It should be noted that the above method of derivation is
entirely general, i.e. it will apply to a point P in any quad
rant and to any angle a.
138. The Addition Formulas. We may now enter upon a
more detailed study of the properties
of the trigonometric functions. We
shall first express sin (a + p) and
cos (a + /8) in terms of sin a, cos a,
sin p, cos 13.* To this end let OP be
the terminal side of any angle a (Fig.
129). If OP is then rotated about
through an angle jS to the position
OP, the terminal line of the angle
a \ p is OP'. If P has the coordinates (x, y) and P the
coordinates {x\ y'), then from (1) § 137,
x' = xcos p — y sin /8,
y' = X sin ^ + y cos p.
Now sin {a + y8) is by definition equal to ^ and cos (a 4 /?)
OP' = OP. Hence
sin(a+ fi)=^ =  sin/3 4^ cos^S,
7 r r
sin (a + P) = sin a cos p + cos a sin p.
T
/
Mv)
/
'' p^
L^
V
W"
X
X
Fig. 129
to — where r
r
or
(1)
Also
or
(2)
cos (a + ;8) =  =  cos ^  ^ sin^,
r r r
cos (a 4 P) = cos a cos p — sin a sin p.
♦We have already had occasion to note that sin (a + ^) is not in general
equal to sin a + sin ^. (See Ex. 5, p. 151.)
202 MATHEMATICAL ANALYSIS [VII, § 138
Further we have
tan (a4 B) = sin (a \ /?) __ sin « cos ^ + cos ce sin yg
cos {a f y8) cos a cos /3 — sin a sin)8*
Dividing numerator and denominator by cos a cos )8, we have
(3) tan(a + 3) = ^^ + ^^ °^.
Furthermore, by replacing yg by — /? in (1), (2), and (3), and
recalling that
sin (—/?)= — sin 13, cos (— ^)= cos /?, tan (— yS) = — tanyS,
we obtain
(4) sin(a— p)=sinacosp— cosasinp,
(5) cos (a — p) = cos a cos p + sin a sin p,
(6) tanCa S)= tangtanp
v; i^ii^a p; i_^tanatanp
EXERCISES
Expand the following :
1. sin (45° + a)= 3. cos (60'' + a) = 5. sin (30°  45°) =
2. tan (30°^)= 4. tan (45° + 60°) = 6. cos ( 180°  45°) =
7. What do the following formulas become if a = yS ?
sin (a + /3) = sin a cos ^3 f cos a sin /3. ^^^ /^ , on _ tan ct + tan j3
sin (a — /3) = sin a cos j3 — cos a sin jS. ' ~ 1  tan a tan /3 *
cos (a + /3) = cos a cos /3  sin a sin ^. . , „. _ tan a — tan/3
lan fcc — p) — ■ ■ — — — — .
cos (a — /3) = cos a cos ^ + sm a sin /3. 1 + tan a tan )3
8. Complete the following formulas :
sin 2 a cos a + cos 2 a sin a = tan 2 « + tan a _
sin 3 a cos a — cos 3 a sin a = 1 — tan 2 a tan a ~
9. Prove sin 75° =: ^^ + \ cos75°=^^:\ tan75° = ^5+^.
2\/2 2V2 V31
10. Given tan a = , sin /3 = ^\, and « and ^ both positive acute angles,
find the value of tan (a f /3) ; sin (a — /3) ; cos(a + /3) ; tan (a  /3).
VII, § 138] TRIGONOMETRIC RELATIONS 203
11. Prove that
(a) cos (60° + a) + sin (80° + «) = cos a.
(b) sin (60° + 6) sin (60°  ^) = sin d.
(c) cos (30° f ^)  cos (30°  d) = ^in d.
(d) cos (45° + d)+ cos (45° 6) = V2 • cos 6.
(e) sin(a + J + sinfa — j = sina.
(/) cos(a + ^) + cos(a^) = V3.
(g) tan (45° + ^) = L±ia!L? . (h) tan (46°  5) = ^ ^ ^^° ^ •
12. By using the functions of 60° and 30° find the value of sin 90° ;
cos 90°.
13. Find in radical form the value of sin 15°; cos 15°; tan 16° J
sin 105° ; cos 105° ; tan 105°.
14. If tan a = I, sin j3 = j\, and a is in the third quadrant while /3 is
in the second, find sin (a ± j3) ; cos (a ± /3) ; tan (a ± /3).
Prove the following identities :
j^g sin (ct + /3) _ tan ct + tan /3 jg sin2cs ^^^ ^ ^ = ejn 3 «
sin (a — j3) tan a — tan /3 ' sec a esc a
j^ tan ot — tan (cc — /3) _ ^^ o 19 (a) sin (180° — &) = sin ^.
l + tanatan(a ^) ~ (&) cos(180°  ^) = cos^.
18. tan(0 ± 45°) + ctn {6 T 45°) = 0. (c) tan (180°  ^) =  tan Q.
20. cos (a + /3) cos (a — /3) = cos^ a — sin^ ^.
21. sin (a + jS) sin (a — /3) = sin2 a — sin2 /3.
22. ctn(«+^) = ^^"""^'^^^ 23. ctn («^) = ^ill^^^^+i .
ctn a + ctn j3 ctn /3  ctn a
24. Prove Arc tan ^ + Arc tan \ = ir/i
[Hint : Let Arc tan l = x and Arc tan 1 = y. Then we wish to prove
X iy = 7r/4, which is true since tan (x + y)= 1.]
25. Prove Arc sin a + Arc cos a =  , if < a < 1.
26. Prove Arc sin j\ + Arc sin f = Arc sin .
27. Prove Arc tan 2 } Arc tan ^ = 7r/2.
28. Prove Arc cos  + Arc cos (— ^j) = Arc cos ( — f f ).
29. Prove Arc tan j% + Arc tan  = Arc tan .
30. Find the value of sin [Arc sin  + Arc ctn ].
31. Find the value of sin [Arc sin a H Arc sin 6] if < a < 1, < 6 < 1.
204 MATHEMATICAL ANALYSIS [VII, § 138
32. Expand sin (x { y \ z) ; cos(x + y + z).
[Hint: a; + 2/ + 2 = (x + ?/) + 2.]
33. The area ^ of a triangle was computed from the formula
A = I ab sin 6. If an error e was made in measuring the angle ^, show that
the corrected area A' is given by the relation A' = A{co8€ + sinectn^).
139. Functions of Double Angles. In this and the follow
ing articles (§§ 139141) we shall derive from the addition
formulas a variety of other relations which are serviceable in
transforming trigonometric expressions. Since the formulas
for sin (a + /3) and cos (a + yS) are true for all angles a and fi,
they will be true when JS = a. Putting /? = a, we obtain
(1) sin 2 a = 2 sin a cos a,
(2) cos 2 a = cos2 a — sin2 a.
Since sin^ a + cos^ a = 1, we have also
(3) cos 2 a = 1  2 sin2 a
(4) = 2 C0S2 a — 1.
Similarly the formula for tan (a + /?) (which is true for all
angles a, ^, and a + /3 which have tangents) becomes, when
)8 = «,
^ ^ 1 — tan^a
which holds for every angle for which both members are
defined.
The above formulas should be learned in words. For ex
ample, formula (1) states that the sine of any angle equals
twice the sine of half the angle times the cosine of half the
angle. Thus
sin 6 a; = 2 sin 3 x cos 3 a?,
2 tan 2 x
tan 4 x =
l~tan22a;'
cos a; =5 cos*?— sin^?.
VII, § 140] TRIGONOMETRIC RELATIONS 205
140. Functions of Half Angles. From (3), § 139, we have
2sin^^ =
; 1 — COS a.
Therefore
• «!=
(6)
:±v'~r"
From (4),
§ 139, we have
Therefore
2cos2^ =
cos =
1 + COS a.
(J)
±V'T"
rormulas (6) and (7) are at once seen to hold for all angles
a. Now, if we divide formula (6) by formula (7), we obtain
(8) ta"i=±Vr
— cos a
j cos a
which is true for all angles a except n • 180°, where n is any
odd integer.
Example. Given sin ^ = — 3/5, cos A negative ; find sin (A/2).
Since the angle A is in the third quadrant, A/2 is in the second or
fourth quadrant, and hence sin (A/2) may be either positive or negative.
Therefore, since cos A=— 4/5, we have
2 Al 2 VlO 10
EXERCISES
Complete the following formulas and state whether they are true for
all angles :
1. sin 2 a = 3. tan 2 a = 6. cos ^ =
2
2. cos2a= (three forms). 4. sini= 6. tan =
2 ' ^
7. In what quadrant is e/2 if 6 is positive, less than 360°, and in the
second quadrant ? third quadrant ? fourth quadrant ?
206 MATHEMATICAL ANALYSIS [VII, § 140
8. Express cos 2 a in terms of cos 4 a.
9. Express sin x in terms of functions of 3 x.
10. Express tan 4 a in terms of tan 2 a.
11. Express tan 4 a in terms of cos 8 a.
12. Express sin x in terms of functions of x/2.
13. Explain why the formulas for sin x and cos x in terms of functions
ot2x have a double sign.
14 From the functions of 30° find those of 60°.
15. From the functions of 60° find those of 30°.
16. From the functions of 30° find those of 16°.
17. From the functions of 15° find those of 7.5°.
18 Find the functions of 2 a if sin a = f and a is in the second
quadrant.
19. Find the functions of a/2 if cos a =— 0.6 and a is in third quad
rant, positive, and less than 360°.
20. Express sin 3 a in terms of sin a. [Hint : 3a = 2afa.]
21. From the value of cos 45° find the functions of 22.5".
22. Given sin a = — and a in the second quadrant. Find the values of
13 ^
(a) sin 2 a. (c) cos 2 a. (e) tan 2 a.
(b) sin". (d) cos. (/) tan.
2 2 2
o
23. If tan 2 a =  find sin a, cos a, tan a if a is an angle in the third
4
quadrant.
Prove the following identities :
24. L±^2S^=ctn«. 27. 1  cos 2 ^ + sin 2 g^ ^^^ ^
sin a 2 1 I cos 2 ^ 4 sin 2 ^
26. fsin^cos^j = 1  sin 5. 28. sin^ + cos^ = ± Vl + sin«.
26. cos2g + cosgH^^^^g 29. sec« + tan a = tanf? + «^
siu2e + sin^ \4 2/
30. 2 Arc cos x = Arc cos (2 a;* — 1) .
31. 2 Arc coso; = Arcsin (2 arvl — a;2).
VII, § 141] TRIGONOMETRIC RELATIONS 207
32. tan [2 Arc tan x] = ^^ . 34. tan [2 Arc sec x] = ± ^ ^
1x^  2  x^
1 — r2
33. cos [2 Arc tan x] = ~ — • 35. cos (2 Arc sin a) = 1  2 a^.
•■ • 1 + x^
Solve the following equations :
36. cos 2 X + 5 sin X = 3. 40. sin^ 2 x — sin^ x = .
37. cos 2 X  sin X = . 41. sin 2 x = 2 cos x.
38. sin 2 X cos X = sin X. 42. 2 sin22x = 1 — cos2x.
39. 2 sin2 x + sin^ 2 x = 2. 43 ctn x — esc 2 x = 1.
44. A flagpole 50 ft. high stands on a tower 49 ft. high. At what dis
tance from the foot of tlie tower will the flagpole and the tower subtend
equal angles ?
45. The dial of a town clock has a diameter of 10 ft. and its center is
100 ft. above the ground. At what distance from the foot of the tower
will the dial be most plainly visible ? [The angle subtended by the dial
must be as large as possible.]
141. Product Formulas. From § 138 we have
sin (a 4 j8) = sin a cos ft \ cos a sin p,
sin (a — ^) = sin a cos ^ — cos a sin y8.
Adding, we get
(1) sin (a f iS) + sin {a — 13)= 2 sin a cos jS.
Subtracting, we have
(2) sin (a f /3) — sin (« — ^) = 2 cos a sin ft.
Now, if we let a + /8 = P and a — fi = Q,
then « = ^«, fi = ^.
Therefore formulas (1) and (2) become
. P 4 O P — O
sm P H sm Q = 2 sm — :r_J!^cos — — ^,
2 2
P 4 O . P— O
Sin P — sm Q = 2 cos ^ ^ sm — — ^.
^ 2 2
Similarly, starting with cos (a + /3) and cos (a — ft) and per
forming the same operations, the following formulas result :
208 MATHEMATICAL ANALYSIS [VII, § 141
cos P + COS Q = 2 COS ^±2 COS ^^=2
2 2
COS P — COS p =  2 sin "j" ^ sin ^ »
2 2
In words :
the sum of two sines =
twice sin (half sum) times cos. (half difference),
the difference of two sines =
twice cos (half sum) times sin (half difference),*
the sum of two cosines =
twice cos (half sum) times cos (half difference),
the difference of two cosines =
minus twice sin (half sum) times sin (half difference).*
Example 1. Prove that
cos 3^ +cosa;^^^^g
sin 3 a; + sin x
for all angles for which both members are defined.
cos 3 a; + cosx _ 2 cos ^ (3 x + a:) cos \ (3 a; — a;) _ cos 2 a; _ ^ « ^
sin 3 X + sin x 2 sin \ (3 x + a:) cos ^ (3 x — x) ~ sin 2 x ~
Example 2. Reduce sin 4 x + cos 2 x to the form of a product.
We may write this as sin 4 x + sin (90° — 2 x), which is equal to
2 ,i„ 4a;490°2» ^„3 4»90° + 2a; ^ ^ ^„ ^^^. + x) COS (3 «  45°).
2 2d
EXERCISES
Reduce to a product :
1. sin 4 ^ — sin 2 ^. 4. cos 2 ^ + sin 2 6. 7. cos 3 x + sin 5 x.
2. cos + cos 3 d. 5. cos 3 ^ — cos 6 6. 8. sin 20° — sin 60°.
3. cos65 + cos2^. 6. sin (x 4 Ax) — sin X.
Show that
9. sin 20° + sin 40° = cos 10°. ^^ sin 15° + sin 75° _ _ ^^ g^o
10. cos 50° + cos 70° = cos 10°. * sin 16°  sin 75°
a. sin 75° sin 15° ^^^3Qo^ 13 sin 3 g sin5g^_ ^^^^^^
cos 75° + cos 15° cos 3 6 — cos 50
* The difference is taken, first angle minus the second.
VII, § 142] TRIGONOMETRIC RELATIONS 209
Prove the following identities :
14. sin 4 ct f sin 3 ct _^^^ce jg sin « + sin /3 __ tan ^ (ce H3)
cos 3 a — cos 4 a 2 ", sina — sin/3~tau ^ (a — ^)
4 e cos a + 2 cos 3 a + cos 5 a cos 3 a
JLO. ■ ■ = •
cos 3 « + 2 cos 5 a + cos 7 a cos 5 a
^rj cosctcos/3 _ tan(« + /3) ^g sin (n  2) g + sin w g _ ^^^ ^
cos a + cos ^ ctn i (a 8) ' cos (n— 2)0  cosn^
Solve the following equations :
19. cos 6 4 cos 5 ^ = cos 3 ^. 22. sin 4 — sin 2 ^ = cos 3 d.
20. sin ^ + sin 5 ^ =: sin 3 d. 23. cos 7 ^ — cos ^ = — sin 4 6.
21. sin 3 ^ + sin 7 ^ = sin 6 6.
142. Law of Tangents. A method for shortening computa
tion will be presented in the next chapter. In applying this
method to the solution of triangles the formulas given below
are valuable. We shall state first the socalled law of tangents:
The difference of two sides of" a triangle is to their sum as the
tangent of half the difference of the opposite angles is to the tan
gent of half their sum.
Proof. a^sin^,
5 sin 5 •
Hence, by proportion, we have
g — 6 _ sin ^ — sin ^
a 4 6 sin ^ f sin 5
But
. ^ . ^ 2cos^i±^sin^:^ tan^^
sm ^  sm J5 2 2 2
sin ^f sin ^ o  A \ B A — B . A
~ 2 sm — cos tan —
2 2
tan^^
Therefore ah^ 2_^
« + «> tan4+§
D
210 MATHEMATICAL ANALYSIS [VII, § 143
143. Angles of a Triangle in Terms of the Sides. Con
^ struct the inscribed circle of the triangle
and denote its radius by r. If the perim
eter a + &4c=2s, then (Fig. 130)
AE = AF=s a,
.Usa^F B BD= BF= sb.
Fig. 130. CD = CE = s  c.
Then tani^ = ^, tani.5 = ^^, taniC=^l,
s—a s — b ^ s—c
where, from § 130,
■ J(^«)(g^)(^c)
' MISCELLANEOUS EXERCISES
1. Reduce to radians 65°, — 135°, — 300°, 20°.
2. Reduce to degrees tt, 3 tt, — 2 tt, 4 tt radians.
3. Find sin (a — /3) and cos (a + ^3) when it is given that a and /3 are
positive and acute and tan a =  and sec p = ^.
4. Find tan (a + /3) and tan (a — /3) when it is given that tan a = ^
and tan /3 = ^,
6. Prove that sin 4 a = 4 sin a cos a — 8 sin^ a cos a.
2
6. Given sin ^ = — ^, and d in the second quadrant. Find sin 2 d,
V5
cos 2 ^, tan 2 ^.
Prove the following identities :
. 7. sin2a = 2iHL^. 8. cos2 « = ^ " ^^"'«
1 f tan2 a 1 + tan'^ a
9. sec2a = _5?^i«_ 10. tan«= "^"^" .
csc2 a — 2 1 + cos 2 a
11. sin (a \ j3) cos /3 — cos (a + p) sin /S = sin a.
12. sin 2 a + sin 2 )3 + sin 2 7 = 4 sin a sin /S sin 7, if a + j9 + 7 = 180°.
1 + tan 
la co8« _ 2
1 — sin a I .. a
VII, § 143] TRIGONOMETRIC RELATIONS 211
14. 1 + tan a tan ^
15.
2
sin^ a + cos^ ct _ 2 — sin 2 ce
sin a + cos a " 2
16. «H?l^ = 2cos2«.
sin 2 a
17. Arc cos f 4 Arc tan f = Arc tan W,
Solve the following equations :
18. cos 2 a = cos^ a.
19. 2 sin a = sin 2 a.
20. cos 2 a + cos a = — 1.
21. sin a + sin 2 a + sin 3 a = 0.
22. sin 2 a — cos 2 a — sin a 4 cos a = 0.
23. Arc tan x + Arc tan (1 — x) = Arc tan .
3
26. Arc tan ?tl + Arc tan ^^ = 180° + Arc tan (  7).
x — \ X
26. Arc sin x + Arc sin = 120°.
2
2 tr
27. Arcsina; + 2 Arccosx = ^•
3
In a right triangle ABC, right angled at (7, prove
28. sin2 ^ = £jZi? . 29. cos^ ^ = .^±i: . 30. *tan ^ "
2 2c 2 2c 2 a + 6
31. Solve for x and ?/ the following equations :
ic sin a + ^ cos a = sin a,
X cos a — ?/ sin a = cos a.
32. Solve for x and y the following equations :
a; cos ^ — y sin ^ = sin ^,
X sin ^ + y cos ^ = cos d.
33. If 2 X is less than 90° and sinx=cos(2 x + 40°), find the value of x.
34. Find « so that the equation x^ + 2 x cos a + 1 = shall have equal
roots,
35. Find a so that the equation 3 x^ + 2 x sec a + 1 = shall have
equal roots.
CHAPTER VIII
THE LOGARITHMIC AND EXPONENTUL FUNCTIONS
144. The Invention of Logarithms. In the last two chap
ters we have had occasion to do a considerable amount of
numerical computation. In spite of the fact that we have
confined these computations to comparatively small numbers
and have had the assistance of tables of squares and square
roots, the calculations have often been laborious.
To carry out by the methods thus far at our disposal the
computations involved in many of the problems of insurance,
engineering, astronomy, etc., would require a prohibitive
amount of labor. That it is now practicable jjo effect such
computations is largely due to the invention ©f logarithms by
John Napier (15501617), Baron of Merchjfeton, in Scotland.
As in the case of many epochmaking inventions, the funda
mental idea of Napier was extraordinarily simple. It may be
explained as follows. Consider the function y = 2". We
readily obtain the following table of corresponding values :
(1)
X
1
2
3
4
6
6
7
8
9
10
11
12
2/ = 2^
2
4
8
16
82
64
128
256
512
1024
2048
4096
Now, since 2" • 2' = 2"+'', it is clear that, if we desire to ob
tain the product of two numbers in the lower line of the table,
we need only add the two corresponding numbers in the upper
line (the exponents), and then find the number in the lower
212
VIII, § 145] EXPONENTS — LOGARITHMS 213
line which corresponds to this sum* For example, to find the
product of 128 x 16, we find from the table that the numbers
corresponding to 128 and 16 are 7 and 4, respectively; the sum
of the last pair is 11 and the number in the lower line corre
sponding to 11 is 2048, which is the product sought. Or again,
to find 4096 ? 512, we find the corresponding exponents 12 and
9 in the table, subtract (12 — 9 = 3), and find the required quo
tient to be 8. How would you justify the latter procedure ?
While the fundamental idea here described is simple, con
siderable insight was required to make the idea practicable.
For, the above table makes possible the finding of the product
of two numbers only when the numbers in question and
their product are to be found in the lower line of the table. In
order to be useful in practical computation it is obviously
necessary to construct a table which will contain every number,
or at least from which the corresponding " exponent " of any
number can easily be obtained either precisely or with a high
degree of approximation. Th^ problem confronting Napier
was to Jill in the gaps in the numbers of the lower line of the
table on p. 212, while preserving the fundamental property of
the table, yIz. that to the product of any two numbers of the lower
line corresponds the sum of the two corresponding numbers of the
upper line.
145. Extension of the Table. An examination of table (1)
reveals the following properties : (a) the values of x form an
arithmetic progression (A.P.), since every number after the
first is obtained by adding 1 to the preceding number ; (6) the
values of y form a geometric progression (G.P.), since every
number after the first is obtained by multiplying the preceding
number by 2. These considerations suggest the possibility of
extending the table in two ways.
214
MATHEMATICAL ANALYSIS [VIII, § 145
5
4
3
2
 1
•
1
2
o
4
5
6
1
7
0.03126
0.0625
0.125
0.25
0.5
1
2
4
8
16
32
64
128
In the first place, we may extend it to the left so as to make
the lower line contain numbers less than 2. To do this, we
need only subtract 1 successively from the numbers of the upper
line and divide by 2 successively the numbers of the lower
line. We then obtain a table extending in both directions :
(2)
This table is still satisfactory. If we desire to multiply 128
by 0.0625, we add the corresponding numbers of the upper line,
namely, 7 and — 4 ; thus we obtain the number 3, which
according to the previous rule should give 128 x 0.0G25 = 8,
which is correct. That the rule still applies may be tested on
other products ; the fact that it does will be proved later.
In the second place we may find new numbers to fill the gaps
in the original table, by inserting arithmetic means between
the successive values of x and geometric means between the
successive values of y. Thus, if we take the following portion
of the preceding table
2
1
1
2
3
4
i
i
1
2 1 4
8
IG
and insert between every two successive numbers of the upper
line their arithmetic, and between every two successive num
bers of the lower line their geometric mean, we obtain the table
(3)
2
f
 1
h
i
1
3
2
5
2
3
1
4
i
^\/2
h
iV2
1
V2
2
2V2
4
4V2
8
8V2
16
VIII, § 145] EXPONENTS — LOGARITHMS
215
If the radicals are expressed approximately as decimals, this
table takes tlie form
2.0
1.5
 1.0
0.5
'
0.5
1.0
1.5
2
2.5
3
3.5
4
0.25
0.35
0.50
0.72
l.OC
1.41
2.00
2.8S
4.00
5.66
8.00
11.31
16
x(AP.)
0.00
0.25
0.50 0.75
1.00
1.25 1.50
1.75
2.00
2.25
yCG.R)
1.00
1.19
1.41 ' 1.68
2.00
2.38 2.83
3.36
4.00 ' 4.76
1
Repeating this process of inserting means, we get the follow
ing table. To save space, we have begun the arithmetic pro
gression with and the geometric progression with 1, and have
not carried the table as far as in the preceding case.
(4)
The rule for multiplying two values of y seems to apply also
to this table, at least approximately. For example, if we apply
the rule to find 3.36 x 1.19, we note that the sum of the cor
responding values of x is 1.75 f 0.25 = 2.00 and conclude that
3.36 X 1.19 = 4.00. Actual multiplication gives 3.36 x 1.19
= 3.9984. The discrepancy we may attribute to the fact that
the values of y other than 1, 2, 4 are only approximations to
the true values.*
The process used in constructing this table may be continued
indefinitely. It enables us to interpolate a new value of x be
tween any two successive values of x and a new value of y
between the two corresponding values of y. But this means
that we can make the values of x and y as dense as we please,
in other words, we can make the difference between successive
values of y as small as we please. By continuing the process
♦ In fact the rules for computing with approximate numbers would lead us
to write 4.00 in place of 3.9984 as we have no right to retain more than two
decimal places. See § 160.
216
MATHEMATICAL ANALYSIS [VIII, § 145
long enough we can make any number appear among the values
of y to as high a degree of approximation as we desire and our
intention of filling the gaps will then be attained. We must
now prove, however, that the rule for multiplication does really
hold in the extended table. Thus far we have merely verified
this rule for special cases.
EXERCISES
1. Assuming that the rule for multiplication applies, find by means of
table (4) the following products.
3.36 X 1.41, 1.68 X 2.38, (1.68)2, (1.19)6.
Check by ordinary multiplication.
146. Arithmetic and Geometric Progressions. The tables
constructed consist of an arithmetic progression one term of
which is the number (the terms of this arithmetic progression
we denoted by x) and a geometric progression one term of
which is the number 1 (the terms of this geometric progression
we denoted by y). Moreover, to every value of x corresponds a
definite value of y in such a way that to oj = corresponds 2/ = 1,
and that to each succeeding (or preceding) value of x corre
sponds the succeeding (or preceding) value of y. Now suppose
that the common difference of the arithmetic progression is d
and that the common ratio of the geometric progression is r.
The correspondence between the values of x and y would then
be exhibited in the following table.
(5)
We shall now prove that in this tahle^ to the product of any
two valiLes of y corresponds the sum of the two corresponding
values ofx.
X
...
md
...
3d
2d
d
d
2d
Sd
...
nd
...
y
...
1
...
1
1
1
r
1
r
r^
r8
r"
...
Vni, § 147] EXPONENTS — LOGARITHMS
217
If tlie two values of y are both, to the right of y = 1, for ex
ample i/i = r^, y.2 = J*^, then the corresponding values of x are
pd and qd. To the product 2/12/2= ^^^'^ corresponds {p\q)d.
If the two values of y are both to the left of 2/ = 1, the proof is
similar. It is left as an exercise.
If one value is to the left of 2/ = 1, for example, y — l/r^, and
the other value is to the right, for example 2/2 = ''^j the cor
responding values of x are — pd and qd respectively. The
product 2/12/2 is equal to (1/r^) r« = r^~^ if q> p, and is equal
to l/r^« if q<p. The value of x corresponding to 2/12/2 is
then {q — p)d/\i q> p and —{p — q)d ii q<,p. But {q — p)d
= — {p —q)d = qd \{— pd). The discussion of the case
p = q\^ left as an exercise. If one of the values of y is 1, the
desired result follows immediately. Why ?
In view of this theorem the validity of the rule used in the
last article for multiplication is established. For tables (2)
and (4) are both tables of the type (5), the former having
d = 1 and r = 2, the latter having d — 0.25 and r = V2 = 1.19
(approximately) .
147. The Exponential Function (i^{a > 0). Let us now con
sider the table
X
— m
...
3
2
 1
1
2
3
...
n
...
y
...
1
a3
1
a2
1
a
1
a
a2
a^
a'»
...
where a represents any positive number.* This table defines y
as a function of x. Morover, this table is a table of the type
(5) ; and all tables obtained by interpolating arithmetic means
between two successive values of x and the same number of
geometric means between the corresponding values of y are of
* The value a = 1 leads to trivial results. Hence, we assume also that a ^ 1.
218
MATHEMATICAL ANALYSIS [VIII, § 147
the type (5). Thus if we interpolate q arithmetic means be
tween x — and x = 1, and q geometric means between y —1
and y= a, we obtain the following table :
X
...
...
1
2
_1
1
Q
2 1 ...
y
...
1
...
1
a
1
1
1
la
^.'/;;^2 ...
V^f
i 1
X
1
9 + 1
Q
...
2
...
.
iv'ar'
a
{Vay^'
...
a2
{VaY
...
which is a table of the type (5), with d = 1/q and r = va.
The function y oi x thus defined is y = a% for ic = 1, 2, 3, •••.
We are therefore led to define the expression a* for fractional
and negative values of x and for x = as follows :
(1) aO = l.
(2) ai/« means ^o", where q^ is a positive integer.
(3) aP/^ means (Va)^, or its equal V^y where p and q
are positive integers.
(4) a^ means 1/a**, where n is any positive rational number.
In view of the fundamental property of any table of type
(5), whereby to the product of any two values of y corresponds
the sum of the two corresponding values of x, we have
(t^ . av — flM+f
for^all values of u and v for which the expressions a", a", and
a""*"' have been defined.
The function ?/ = a* (a > 0) has now been defined for all
rational values of x. To complete the definition of this func
* We should keep in mind that the symbol ^a (o>0) means the positive
gth root of a. Thus yfm = 2, not — 2.
VIII, §147] EXPONENTS — LOGARITHMS 219
tion for all real values of a;, we must indicate the mean
ing of a* when x is an irrational number. To carry this
definition through in all its details is beyond the scope of
an elementary course. But we have seen that any irrational
number may be represented approximately by a rational num
ber, with an error as small as we please. (See § 29.) Thus
V3 is represented approximately by the rational numbers 1.7,
1.73, 1.732, ... Our previous definitions have given a definite
meaning, for example, to 2^^, 2^''^, 2'^''^^, .... The values of the
latter expressions are by definition approximate values of 2^^.
We take for granted without proof the fact that the successive
numbers
(6) 2^'\ 217^ 21"^, ...,
as the exponents represent closer and closer approxima
tions to V3, approach closer and closer to a definite number.
This definite number is by definition the value of 2^. Similar
considerations apply to the definition of a^, where a is any
positive number and x is any irrational number. The principle
involved is briefly expressed as follows :
An approximate value of x gives an approximate value of a'.
The value of a* can he found as accurately as we please by using
a sufficiently accurate approximation to x.
The objection might be raised that the calculation of '2?''^ involves the
extraction of the 10th root of 2 and the calculation of 2^'^ involves the
extraction of the 100th root of 2, etc., and perhaps we do not know how
to extract these roots. As a matter of fact we can calculate 2^^ as ac
curately as we please by extracting square roots only. The processus as
follows : We know that \/3 = 1.7320 accurately to four decimal places.
Now by table (4), p. 215, we see that 2^^2M and 2'" = 3.36. We carry
the computation to more places and have 2i«)oo =2.8284 and 2i7500= 3.3636.
Now, 1.7320 lies between 1.5000 and 1.7500, the arithmetic mean of
which is 1.6250. The geometric mean of 2.8384 and 3 3635 is 3.0844
According to our previous definitions we have then 2i<'25o = 3.0844.
220 MATHEMATICAL ANALYSIS [VIH, § 147
Inserting means between the last two results we have 2^^^ = 3.2200.
By inserting arithmetic means between the properly selected exponents
and geometric means between the corresponding powers of 2 we can ulti
mately obtain the value of 2i^320, ^he results of the necessary steps are :
21.7188 :^ 3.2915, 217344 = 3.3274, 217266 = 3.3094, 2i7305 = 3.3182,
21.7325 == 3.3228, 217315 = 3.3205, 2i7320 = 3. 32 17.
The process here illustrated makes it possible to calculate 2^3 to as high
a degree of approximation as we please, since we can carry the computa
tion to as large a number of decimal places as we please.
148. The Laws of Exponents. The function y = a'' (a > 0)
is now defined for all real values of x. This function is called
the exponential function of base a. The laws of exponents
I. a^ ' a^ = a^^^ ]
II. (cr*)'' = a«*» , a > 0, 6 > 0,
III. a«* . b^ = (ab)^ '
which were derived previously (§ 42) for positive integral ex
ponents, hold for all real values of u and v. The first of these
we have already derived. The last two may be readily proved
for negative, fractional, and zero exponents by using the defini
tion of a''.
Thus by definition, if it = p/q and v = 71, where p, q, n are
positive integers, we have
p pn
If u is any positive rational number and v=p/qf where p, q are
positive integers, we have,
up
(a^y = (a«)« = Via'^y = ^/a"^ = a «" = a"".
If u = — n, where n is a positive rational number, and if
V =p/q, where p and q are positive integers, we have
(ay = (a«)^/'=/^\/iY= — i— = i = a" 7 = a«^
If u is any rational number and v =— n, where n is any
VIII, § 149] EXPONENTS — LOGARITHMS
positive rational number,
221
(a«)''=(a'*)"
— — — = — = a""" = a*
If either w or v is zero, the result is immediate. Hence the law
II is proved for rational exponents.
A similar proof of the law III is left as an exercise.
149. The Graph of the Exponential Function. Figure 131
represents the graph of the function y = 2*, drawn from the
tabular representation given in the first table on p. 215.
~

7
r
^
~
n
\
l.
1
\
u
\
[e
ri//^/^
h
12
\
R

«l \\
.U
's
i
\
i
/
\
/
/
'
/
/
?^
/
y
j
/
"^
/
1
/
r
\
'
/
/
/
\\
/
/
/
/
<^
\ A
\
/
y
<\
/
' 1
•
'5'
N
s^
V.
/
X
"*"
1
y
^
^
^
a
= 1
^
1
^
^
?J
,
_
—J
4
'
V
=2tc
V
=
^
_
Fig. 131
FiQ. 132
It will be noted that all curves of the system ?/ = a* pass
through the point (0, 1). By hypothesis a > 0. If a > 1, the
function a^ is an increasing function ; while if a < 1, the func
tion is a decreasing function. Figure 132 shows some of the
curves of the system y = a*.
222 MATHEMATICAL ANALYSIS [VIII, § 149
EXERCISES
1. Calculate the value of the exponential function 3*
(a) for the values of a; = 1, 2, 3, 4, 0, — 1, — 2,  3, — 4 ;
(b) for the values x = 0.5, 1.5, 2.5, 3.5,  0.5,  1.5,  2.5 ;
(c) for the values x = 0.25, 0.75, 1.25, 1.75.
Arrange the results in the form of a single table.
2. Show how to use the table constructed in Ex. 1 to solve problems in
multiplication, division, raising to powers, and extracting roots. Make up
your own problems and check your results by the methods of arithmetic.
3. Describe in detail how you would find the value of 3 . Between
what two numbers in the table found in Ex. 1 does the value of 3^^ lie ?
4. Construct the graph of y = 3* for values of x between — 2 and 3.
6. What is meant by a^ ? x^ ? (l/y)^ ?
6. What is the value of 8^ ? 27^ ? (0.001)^ ? (i)8 ?
7. Simplify (18)^ i (3)^'
8. Perform the following indicated operations :
(«) (^¥. w (^>^K
\x!^y~^y
(6)(aWc¥. (,)^^_,+ 2,.i)..
(c) (32x02/10)^. (/) (2^)*.
9. Multiply
(a) {a^ + a)ia^a). (6) (aiao)(a2a'0(a»aO
(c) (a;^y^)(a;« y^).
(d) (xi 4 x~^y~^ + jr^Xxi  x~^y~^ + y^).
8 2 12 1
(e) (a*  2 a* + 3 a*) (2 a' a* + 2).
8 2 12 1
(/) (ir  oir + 3 6y«  c) (jr + 6y«  cyO).
10. Divide
(a) (a;+l)by (v^ + 1). (&) (aj^'  y*) by (x^ yi^).
(c) (J  a6^ + ah  b^) by (a^  6^).
(d) (ai + 4 a^ + 6 a^ + 4 a^ + a*) by (a"^ + a).
VIII, § 150] EXPONENTS — LOGARITHMS 223
11. Simplify
(a) 12" + I  9' + ^ + 27*. (6) ( '"'P ]'*.
^64 m^p^
12. Simplify
13. Which of the two numbers V5 and ^/8 is the greater and why ?
14. Simplify
(2^ x2^)f(54)i
15. Prove that, if
_ 1 Tx^ _ x"^"
2Lj „d'
■y y
then
2vxy
16. Reduce to simplest form
(c) (a^ + X*) (a2 _ x^)~^  (a2 _ x^)^.
150. Definition of the Logarithm. The logarithm of a
number JV to a base 6 (6 > 0, ^^t: 1) is the exponent x of
the power to which the base h must be raised to produce the
number N.
That is, if
then
X — logbN.
These two equations are of the highest importance in all work
concerning logarithms. One should keep in mind the fact
that if either of them is given, the other may always be
inferred.
224
MATHEMATICAL ANALYSIS [VIII § 150
Tlie graph of the logarithm function (Fig. 133) is obtained
from the graph of the corresponding exponential function by
simply turning the latter graph over about the line through
the origin bisecting the first and third quadrants.
Y

,
<
.— ■
*
^
^
^
x"
'"
/
^
/
f
1 J
1
>
1
s
^^x\
■ 1
V
= z
ff^
X
.
_
_
_J
_
Fig. 133
EXERCISES
1. When 3 is the base what are the logarithms of 9, 27, 3, 1, 81, ^,
2. Why cannot 1 be used as the base of a system of logarithms ?
3. When 10 is the base what are the logarithms of 1, 10, 100, 1000 ?
4. Find the values of x which will satisfy each of the following
equalities :
(a) logs 27 = X. (d) loga a = x. (g) logs x = 6.
(6) log^3 = 1. (e) logal = x. {h) logssx = i
(c) log, 6 = i. (/) logaij^r = «• (0 log.ooi x = .00001.
6, Find the value of each of the following expressions :
(a) loga 16. (c) logesig. (e) log26l26.
(6) log34a49. (d)log2Vl6. (/) log2^.
VIII, § 151] EXPONENTS — LOGARITHMS 225
151. The Three Fundamental Laws of Logarithms. From
the properties of tlie exponential function (p. 220) we derive
the following fundamental laws.
I. Tlie logarithm of a product equals the sum of the logarithms
of its factors. Symbolically,
lege, MN = logft Af + logft N.
Proof. Let logj, M= x, then 6* = M. Let log^ N~y, then
b"=N. Hence we have MN= ft*"^", or
logj,i¥iV=a; + 2/, i.e. \o^^,MN=\og^,M+\o^^N.
II. The logarithm of a quotient equals the logarithm of the
dividend minus the logarithm of the divisor. Symbolically
M
^^Sft TT = logft M — logft N.
N
Proof. L^t logj, Jf = x, then b' = M. Let logj, ^ = y, then
h" = N. Hence we have M/N = &*"»', or
M M
^og,—=xy, i.e. log, — = log,Jflog,JV
IIL T%e logarithm of the pth power of a number equals p
times the logarithm of the number. Symbolically
log6MP = />log5M.
Proof. Let log^ M^x, then 6^ = M. Raising both sides
to the pih. power, we have b^ = M^. Therefore
logj, M^ =px=p log, M.
From law III it follows that the logarithm of the real positive
nth root of a number is one nth of the logarithm of the number.
Q
226 MATHEMATICAL ANALYSIS [VIII, § 151
EXERCISES
1. Given logio 2 = 0.3010, logio 3 = 0.4771, logio 7 = 0.8451, find the
value of each of the following expressions:
(a) logio 6. (/) logio 6.
[Hint: logio 2 X 3=logio 2+logio3.] [Hint : logio5 = logio V]
(6) logio 21.0. (.9) logio 150.
(c) logio 20.0. W logio Vli.
{d) logio 0.03. (i) logio 49.
(«) logio 1. U) logio V2V7^.
2. Given the same three logarithms as in Ex. 1, find the value of each
of the following expressions:
f^\ i«„ 4 x5 x7 ., . i^„ 5 X 3 X 20 ,. , ^ 2058
{d) logio (2)25. (e) logio (3)8(5)6. (/) logio (28)().
152. The Systems most Frequently Used. From the defi
nition of a logarithm (§ 150) any positive number except 1 can
be used as the base of a system of logarithms. As a matter of
fact, however, the numbers generally used are (1) a certain
irrational number which is approximately equal to 2.71828
and is denoted by e and (2) the number 10. Logarithms to the
base 6 are important in certain theoretical problems ; loga
rithms to this base are called natural. For numerical compu
tation it will be seen presently that the base 10 has numerous
advantages. Since different systems of logarithms are in use,
it is important to know how to change from one system to
another. The following law explains how this can be done.
IV. The logarithm of a number M to the base b is equal to the
logarithm of M to any base a, divided by the logarithm of b to the
ba^e a. Symbolically,
lOga O
VIII, § 153] EXPONENTS — LOGARITHMS 227
Proof. Let logj, M=x, then ¥ — M. Taking the logarithms
of both sides to the base a, we have
log^ ?>=^ = log„ 3f, or a; log„ 6 = log, 3f,
log« h
153. Logarithms to the Base 10. Logarithms to the base
10 are known as common logarithms, or as Briggian logarithms,
after Henry Briggs (15561631) who called attention to the
advantages of 10 as a base. These advantages appear below.
If 10 is the base, log 10 = 1, log 100 = 2, log 1000 = 3, etc.
It follows that if a number be multiplied by 10, or by any
positive integral power of 10, the logarithm of the number is
increased by an integer. In other words, the shifting to the
right of the decimal point in a number changes only the in
tegral part of the logarithm and leaves unchanged the decimal
part of the logarithm.
An example will make this clear. Given logio 2 = 0.30103, we have
logio 20 = 1.30103, logio200 = 2.30103, etc. Or, again, given logio 4.5607
= 0.65903, we have logio 45.607 = 1.65903, logio 456.07 = 2.65903,
logio 4560.7 = 3.65903, logio 45607.0 = 4.65903.
It should be clear from these examples that the decimal part
of the logarithm of a number greater than 1 in this system
depends only on the succession of figures composing the num
ber, irrespective of where the decimal point is located ; while
the integral part of the logarithm of the number depends
simply on the position of the decimal point.
The decimal part of a logarithm is called its mantissOj the
integral part its characteristic. In view of what has been said
above only the mantissas of logarithms to the base 10 need be
tabulated. The characteristic can be found by inspection.
This follows from the following considerations.
228 MATHEMATICAL ANALYSIS [VIII, § 153
The common logarithm of a number between 1 and 10 lies
between and 1.
The common logarithm of a number between 10 and 100 lies
between 1 and 2.
The common logarithm of a number between 100 and 1000
lies between 2 and 3.
The common logarithm of a number between 10" and lO'*'^^
lies between n and n + 1.
It follows that a number with one digit (^0) at the left of
the decimal point has for its logarithm a number equal to + a
decimal ; a number with two digits at the left of its decimal
point has for its logarithm a number equal to 1 + a decimal ; a
number with three digits at the left of the decimal point has
for its logarithm a number equal to2f a decimal, etc. We
conclude, therefore, that the characteristic of the common loga
rithm of a number greater than 1 is one less than the number of
digits at the left of the decimal point.
Thus, as before, logio 456.07 = 2.65903.
The case of a logarithm of a number less than 1 requires
special consideration. Taking the numerical example first con
sidered above, if logjo 2 =0.30103, we have logio 0.2 = 0.301031.
Why ? This is a negative number, as it should be (since the
logarithms of numbers less than 1 are all negative, if the
base is greater than 1). But, if we were to carry out this
subtraction and write logjo 0.2 = — .69897 (which would be
correct)^ it would change the mantissa, which is inconvenient.
Hence it is customary to write such a logarithm in the form
9.30103  10.
If there are n ciphers immediately following the decimal
point in a number less than 1, the characteristic is — n — 1.
For convenience J if n < 10, we write this as (9 — n) — 10. This
Vni, § 154] EXPONENTS — LOGARITHMS 229
characteristic is written in two parts. The first part 9 — n is
written at the left of the mantissa and the — 10 at the right.
In tlie sequel, unless the contrary is specifically stated we
shall assume that all logarithms are to the base 10. We may
accordingly omit writing the base in the symbol log when there
is no danger of confusion. Thus, the equation log 2 = 0.30103
means logjo 2 = 0.30103.
154. Use of Tables. Since the characteristic of the loga
rithm of a number may be found by inspection, a table of
logarithms contains only the mantissas. To make practical
use of logarithms in computation it is necessary to have a con
veniently arranged table from which we can find (a) the
logarithm of any given number, and (6) the number corre
sponding to a given logarithm. Tables of logarithms differ
according to the number of decimal places to which the man
tissas are given and also in incidental details. However, the
general principles governing their use are the same. These
principles are explained for a fourplace table (p. 536) by the
following examples.
Problem A. To find the logarithm of a given number,
(1) When the number contains three or fewer figures.
Example. To find the logarithm of 42.7.
First, by §153, the characteristic is 1. We accordingly write (provi
^'°°^'y^ log42.7 = l.
Next we look up in the tables the mantissa corresponding to the succes
sion of figures 4, 2, 7. We run a finger down the first column of the
table until we reach the figures 4, 2, hold it there while with another
finger we mark the column headed with the third figure, 7. At the
intersection of the line and column thus marked, we find the desired
mantissa : 6304, The desired result is then
log 42.7 = 1.6304.
230 MATHEMATICAL ANALYSIS [VIII, § 154
To find the logarithm of 0.0427, we should proceed in precisely the
same way, the only difference being that the characteristic is now 8 — 10.
H®^<^®' log 0.0427 = 8.6304  10.
(2) When the number contains four significant figures
Example. To find log 32.73.
We see that again the characteristic is 1, and we write provisionally
log32.73 = l.
Now, the mantissa of log 32.73 lies between the mantissas of log 32. 70 and
log 32.80; i.e. (from the table) between 5145 and 5159. The difference
between these two mantissas (called the tabular difference at that place in
the table) is 14, und this difference corresponds to a difference in the
numbers of .10. According to the principle of linear interpolation,* the
difference in the mantissas corresponding to a difference in the numbers
of .03 is 14 X .3 = 4.2 or (rounded) 4. The mantissa corresponding to
3273 is then 5145 + 4 = 6149, and we obtain
log 32.73 = 1.5149.
Problem B. To find the number corresponding to a given
logarithm. Here we simply reverse the preceding process.
Example. To find the number whose logarithm is 0.8485.
We first seek the mantissa 8485 in the table. We find that it lies be
tween 8482 and 8488, corresponding respectively to the successions of
figures 7050 and 7060. The tabular difference here is 6, while our differ
ence, i.e. the difference we have to account for (8485 — 8482) is 3.
Hence the corresponding difference in the numbers is  of 10 or 6. Hence
the succession of figures in the number sought is 7055. Since the char
acteristic is 0, the number sought is 7.055. Or, log 7.055 = 0.8485.
If the mantissa is found exactly in the table, of course no interpolation
is necessary. Thus the number whose logarithm is 9.7348 — 10 is 0.5430.
EXERCISES
1. Find the logarithms of the following numbers from the table on
pp. 5367 : 482, 26.4, 6.857, 9001, 0.5932, 0.08628, 0.00038.
2. Find the numbers corresponding to the following logarithms :
2.7935, 0.3502, 7.9699  10, 9.5300 10, 3.6698, 1.0958.
* One should convince oneself that the conditions for linear interpolatiou
are satisfied by this table. In fact, it is readily seen that for several numbers
immediately preceding and following 327, the tabular differences are 13 and 14.
VIII, § 155] EXPONENTS — LOGARITHMS 231
155. Use of Logarithms in Computation. The way in which
logarithms may be used in computation will be sufficiently ex
plained in the following examples. A few devices often neces
sary or at least desirable will be introduced. The latter are
usually selfexplanatory. Reference is made to them here, in
order that one may be sure to note them when they arise.
The use of logarithms in computation depends, of course, on
the fundamental properties derived in § 151.
Example 1. Find the value of 73.26 x 8.914 x 0.9214.
We find the logarithms of the factors, add them, and then find the
number corresponding to this logarithm. The work may be arranged as
follows :
Numbers
Logarithms
73.26
(>)
1.8649
8.914
(»
0.9501
0.9214
(»
9.9645  10
12.779510
601.9 ^ws. '
(^)
2.7795
Product
Example 2. Find the value of 732.6 ^ 89.14.
Numbers Logarithms
732.6 (^^) 2.8649
89.14 (>) 1.9501
Quotient = 8.219 Arts. (<) 0.9148
Example 3. Find the value of 89.14  732.6.
Numbers Logarithms
89.14 (>>) 11.9601  10
732.6 (^) 2.8649
Quotient = 0.1217 Ans, (<~) 9.0852  10
Example 4. Find the value of 763.2 x 21.63
Whenever an example involves several different operations on the
logarithms as in this case, it is desirable to make out a blank form. When
a blank form is used, all logarithms should be looked up first and entered
in their proper places. After this has been done, the necessary opera
tions (addition, subtraction, etc.) are performed. Such a procedure
saves time and minimizes the chance of error.
232
MATHEMATICAL ANALYSIS [VIII, § 155
Form
Numbers
Logarithms
763.2
(»
21.63
(» (+).....
product
986.7
Ans.
(» () ." '.'.'.
....
(<)
Form Filled In
Numbers
Logarithms
763.2
(>) 2.8826
21.63
(>) 1.3351
product
4.2177
986.7
(>) 2.9942
16.73
Ans.
i<) 1.2235
Example 5. Find (1.357)6.
Numbers
Logarithms
1.357
(>) 0.1326
(1.357)6 :
= 4.602
Ans. (^) 0.6630
Example 6. Find the cube root of 30.11.
Numbers
Logarithms
30.11
(_>) 1.4787
y/SOAl =
3.111
Ans. «) 0.4929
Example 7. Find the cube root of 0.08244.
Numbers Logarithms
0.08244 (>) 28.916130
v^O.08244 = 0.4352 Ans. (^) 9.6387  10
EXERCISES
Compute the value of each of the following expressions using the table
on pp. 536537.
1. 34.96 X 4.65. 5. (34.16 x .238)2.
2. 518.7 X 9.02 x .0472. 6. 8.572 x 1.973 x (.8723)8.
„ 0.5683
0.3216
6.007 X 2.483
6.524 X LUO'
7.
648.8
^(21.4)2
\2791
Vm, § 155] EXPONENTS — LOGARITHMS 233
9.
10.
11.
J 2.8076X 3.184
^ (2.012)3
»/ 2941 X 17^32
'>'2173 X 18.76*
\/0. 00732
V735 ^^
12, (20.027)i. (2.01)i
17. The stretch s of a brass wire when a weight m is hung at its free
end is given by the formula ,
where m is the weight applied in grams, g = 980, I is the length of the
wire in centimeters, r is the radius of the wire in centimeters, and yfc is a
constant. If m = 844.9 grams, I — 200.9 centimeters, r = 0.30 centi
meters when s = 0.056, find k.
18. The crushing weight P in pounds of a wrought iron column is given
by the formula ^ 55
P = 299,600^^,
where d is the diameter in inches and I is the length in feet. What weight
will crush a wrought iron column 10 feet long and 2.7 inches in diameter ?
19. The number n of vibrations per second made by a stretched string
is given by the relation . , —
2lM m'
where I is the length of the string in centimeters, ilf is the weight in grams
that stretches the string, m the weight in grams of one centimeter of the
string, and g =980. Find n when i{f= 5467.9 grams, Z = 78.5 centi
meters, m = 0.0065 grams.
20. The time t of oscillation of a pendulum of length I centimeters is
given by the formula
t = rJL.
>'980
Find the time of oscillation of a pendulum 73.27 centimeters in length.
21. The weight w in grams of a cubic meter of aqueous vapor saturated
at 17° C. is given by the formula
^ ^ 1293 X 12.7 X 5
computer. (1+M)(760x8)
234 MATHEMATICAL ANALYSIS [VIII, § 156
156. Exponential Equations. An equation in which the
unknown is contained in an exponent is known as an exponen
tial equation. Some such equations may be solved by taking
the logarithms of both sides after the equation has been
properly transformed.
2x41
Example 1. Solve the equation 3 + 7 = 15.
Transposing the 7 and taking logarithms of both sides we obtain
{2x + l)log3 = log8.
Hence we find
^^iH^^ij.
2Llog3
Example 2. Money is placed at interest, compounded annually. Find
a formula for the amount at the end of n years. Also a formula* giving
the number of years necessary to produce a given amount.
Let C be the original capital and r the given rate of interest {i.e. if
the interest is 5 per cent, r = 0.05). The amount A\ at the end of the
first year is ^ ^ ^ ^,,
Ai= G+ Cr= C{\ + r).
At the end of two years we have
^2 = ^1(1 + = 0(1 + 02.
At the end of n years, the amount is
A=A,= G{\^rY.
This is the formula required. To find w, given A, O, r, we take the
logarithms of both sides and find
EXERCISES
1. Solve for x the equation 2^ = 5.
2. Solve for y the equation 3w + 2 = 9.
■ 3. Solve for x and y the simultaneous equations S^+v = 4, 2*~v = 3.
4. Solve for x and y the simultaneous equations 2*+f = 6y, 3*^ = 2^+^
6. Find the amount of $1000 in 25 years at 6 per cent compound
interest, compounded annually.
VIII, § 1 56j EXPONENTS — LOGARITHMS  235
^ 6. Find the amount of $ 600 in 10 years at 4 per cent compound inter
est, compounded semiannually.
7. In how many years will a sum of money double itself at 5 per cent
interest compounded annually ? semiannually?
8, A thermometer bulb at a temperature of 20° C. is exposed to the air
for 15 seconds, in which time the temperature drops. 4 degrees. If the
law of cooling is given by the formula 6 = ^oe~^S where is the final tem
perature, do the initial temperature, e the natural base of logarithms, and
t the time in seconds, find the value of b.
MISCELLANEOUS EXERCISES
1. "What objections are there to the use of a negative number as the
base of a system of logaiithms ?
2. Show that a^^g^"^ = x.
3. Write each of the following expressions as a single term :
(a) log x\\ogy — log z. (6) 3 log a; — 2 log y + 3 log z.
(c) 3 log a  log (x + y) I log (ex + d)+ log VwT~x.
4. Solve for x the following equations :
(a) 2 log2 X h log2 4 = 1. (c) 2 logio x — 3 logio 2 = 4.
(b) logs «  3 logs 2=4. {d) 3 logs x + 2 loga 3=1.
5. How many digits are there in 235 ? 3142 ? 312 ^28?
6. Which is the greater, (il)^^^ or 100 ?
7. Find the value of each of the following expressions. (See § 152. )
(a) logeSS. (6) logs 34. (c) log7 246. (cZ) logi3 26.
8. Prove that logb a • logo b = 1.
9. Prove that
log, x{Vx^l ^ 2 log, lx+ v^^^l].
X — y/x'^ — 1
10. The velocity v in feet per second of a body that has fallen s feet is
given by the formula v = \/64.3s.
What is the velocity acquired by the body if it falls 45 ft. 7 in. ?
11. Solve for x and y the equations : 2^ = IGv, aj + 4 y = 4.
CHAPTER IX
NUMERICAL COMPUTATION
I. ERRORS IN COMPUTATION
157. Absolute and Relative Errors. In § 29 we noted that
the numerical result of every observed measurement is an
approximation. The difference between the exact value of
the magnitude and this observed value is a concrete number
called the absolute error* Often the absolute error is not the
most serviceable measure of the precision of a measurement.
The relative error, which is defined as the ratio of the absolute
error to the exact value, is often found more serviceable. Since
the relative error is a ratio, it is an abstract number, and is
therefore sometimes expressed in per cent. For example, if
the diagonal of a square 10 in. on a side be measured and
and found to be 14.1 in., the absolute error is less than 1/10
of an inch. The relative error is less than (1/10) ^ 10 \/2
= 1/141, approximately, le. less than 0.71 per cent.
158. Rounded Numbers. Significant Figures. When the
result of a measurement is expressed in the decimal notation,
a generally adopted convention makes it possible to determine
the degree of precision of the measurement from the number
of significant figures contained in the number expressing the
measure. This convention simply specifies that no more digits
shall be written than are (probably) correct. Thus a measure
* The absolute error is therefore positive or negative according as the ob
served value is too small or too large.
236
IX, § 158] NUMERICAL COMPUTATION 237
ment of a length, expressed as 14.1 in. means that the measure
is exact to the nearest tenth of an inch. If on the other hand
the measurefment of this length were exact to the nearest
hundredth of an inch, the measure would have been expressed
by the number 14.10.*
We should note, then, that the two numbers 14.1 and 14.10
do not mean precisely the same thing, when they express the
result of a measurement.
Again we may note that the absolute errors involved in the
expression 4371.52 ft. and 42.81 ft. are each less than one
hundredth of a foot ; whereas the relative error is in the first
case less than 1/437152 and in the second only less than
1/4281.
Sometimes we are furnished with numbers expressing meas
ures which are given with greater accuracy than we can use, or
care to use. Thus suppose we want to express a measured
length of 3.5 in. in terms of centimeters. We find in a table
of equivalent lengths that 1 in. = 2.54001 cm. It would be
obviously absurd to use this expression as it stands. We
accordingly round it off to 2.54 or even to 2.5 and find that 3,5
in. = 8.9 cm. If, on the other hand, we wish to express 3.50000
in. in centimeters, we should have to use the value 2.54001.
A number is rounded off by dropping one or more digits at
the right, and, if the last digit dropped is 5% 6, 7, 8, or 9, in
creasing the preceding digit by l.f Thus the successive
approximations to tt obtained by rounding off 3.14159 . are
3.1416, 3.142, 3.14, 3.1, 3.
* In other words x = 14.1 means that the exact value of x lies between 14.05
and 14.15; and x = 14.10 means that the exact value of x lies between 14.095
and 14.105.
t In rounding off a 6, computers use the following rule : Always round off
a 6 to an even digit. Thus 1.415 would be rounded to 1.42, whereas 1.445
would be rounded to 1.44. The reason for this rule is that, if used con
sistently, the errors made will in the long run compensate each other.
238 MATHEMATICAL ANALYSIS [IX, § 158
The significant figures of a number may now be defined as
the digits 1, 2, 3, , 9 together with such zeros as occur
between them or as have been properly retained in rounding
them off. Thus 34.96 and 3,496,000 are both numbers of four
significant figures. On the other hand 3,496,000.0 has eight
significant figures, since the in the first decimal place accord
ing to the convention adopted means that the number is exact
to the nearest tenth. This zero is then essentially a digit
properly retained in rounding off, and should be counted as one
of the significant figures.
Confusion can arise in only one case. For example, if the
number 3999.7 were rounded by dropping the 7, we should
write it as 4000 which, according to the rule just given, we
would consider as having only one significant figure, whereas in
reality we know from the way in which the number was ob
tained that all four of the figures are significant. When such
a case arises in practice we may simply remember the fact, or
we can indicate that the zeros are significant by underscoring
them, or by some other device. Computers adopt devices of
their own to avoid errors in such cases.
159. Computation with Rounded Numbers. Addition.
Since the (absolute) error of any approximate number can be
at most one half the unit represented by the last digit at the
right, the sum of n such numbers can be in error by at most
n/2 times the unit represented by the last figure. These con
siderations lead to the following convention : in adding a
column of approximate numbers first round off the given
numbers so that not more than one column at the right is
broken ; round off the sum so that the last figure to the right
comes in the last unbroken column. This last figure is then
uncertain. Nevertheless, it is usually retained temporarily.
As a matter of fact, even the figure preceding this last one is
IX, § 160] NUMERICAL COMPUTATION 239
not certain, since the errors may accumulate in adding several
numbers.
Eor example, in adding 21.64
3.8576
5.259743
10.31
we first round off : 21.64
3.858
5.260
10.31
41.068 = 41.07
The final sum is written 41.07, but even the last figure 7 is
open to question. Show that the true result may be as low as
41.06 or as high as 41 .08.
To retain all the figures in the second and third of the num
bers originally given would be absurd and would give in the
result a misleading pretense of accuracy which does not exist
in fact.
In subtraction round off similarly.
160. Multiplication. Let a and b be approximate numbers
and let their relative errors be a and ft respectively. The
exact numbers are then (nearly) a { aa and b + bft. Their
product is a6 + «6«+a&;8 + aJ«j8.
The error committed in using ab as the product is then
ab{a + ^ + «^)
and the relative error is therefore nearly
Now in practice a and /3 are small fractions, so that a ft is in
significant when compared with a + /8. (For example, if a and
240 MATHEMATICAL ANALYSIS [IX, § 160
/3 are both equal approximately to 0.001, ap is equal approxi
mately to 0.000001.) Hence, we conclude that the relative
error of the product of two numbers is approximately equal to the
sum of the relative errors of the factors.
Hence, in finding the product of two approximate numbers,
round off so that the two numbers have the same number of
significant figures, and retain only this number of significant
figures in the product* Even then the last figure retained may
be unreliable.
Example. Multiply 27.17 by^3, 14159. Round off the second factor to
3.142, and multiply:
27.17 X 3.142
5434
10868
2717
8151
85.36814 = 85.37
Even the figure 7 may be in error. Show that the true answer may be as
low as 85.35.
The labor involved in such a multiplication may be considerably re
duced by slightly modifying the method used, as follows :
After having equalized the number of significant fig
ures annex a zero to the multiplicand. Multiply by the
first figure on the left of the multiplier. Drop the last
figure of the multiplicand and multiply by the second
figure of the multiplier. Drop the next figure of the
multiplicand and multiply by the third figure of the
multiplier (but "carry" the amount from the figure
dropped : thus .in the example having dropped the 7 and multiplying by 4,
we say 4 x 7 = 28, carry 3, 4 x 1 = 4, +3 = 7, which is the first figure
we write), and so on, arranging all the partial products so that the last
figures from the left fail into the same vertical column ; then add in the
usual way.
* Since ^^ = 3.1428571, while tt = 3.1415927, the value y may be used for ir
when the uncertainty of the other factors in a product in which it appears
is greater than 1 part in 3000 (approximately) .
27.170 X
3.142
81510
2717
1087
64
85368 =
: 85.37
IX, § 161] NUMERICAL COMPUTATION 241
161. Division. In case either the dividend (iV) or the
divisor (D) is an approximate number, the following shortened
method may be used :
1. Equalize the relative accuracy of N and D ; but if D is
larger at the left, keep one extra figure on JV (as in the example
below).
2. Divide as in long division, but drop successive figures in
Z>, instead of adding successive zeros to N.
Example. Find 295.679 r 7.53. (As 7 is greater than 2, we retain
four figures in the dividend.)
7.53 1 295.7 [39.3
225 9 [3 X 753]
69 8 [divide by 75, gives 9]
67 8 [9x3 = 27, carry 3 ; 9 x 75 = 675, + 3 = 678]
2 [divide by 7, gives 3 (nearer than 2)]
EXERCISES
1. Add the following numbers, each representing the result of a meas
urement: 25.62, 341.718, 2.62394, 28.7125.
2. Express 5.216 inches in centimeters.
3. Express 53.291 cm. in inches.
4. A rectangular table top is measured, and is found to be
2'4".5 X 3'6".4. Find its area. Find the error caused in this area if the
measurements are each O'M too short. Find the relative error in the
area.
5. Assuming that you can estimate the length and the breadth of a
room which is about 15' by 18' to within 2', how nearly can you estimate
its area ?
6. Assuming that you can measure each of the dimensions of the room
of Ex. 5 with a yardstick to within 1" error, how nearly can you find the
area of the floor ? If the height of the room is about 10', how nearly can
you find the volume of the room by measurement ?
7. Assuming that you can measure the radius of a circle about 6" in
diameter to within 0".l error, how nearly can you find its area ? How
nearly could you find by measurement the volume of a cylinder about 5'
high and about 5" in diameter ?
242 MATHEMATICAL ANALYSIS [IX, § 162
II. LOGARITHMIC SOLUTION OF TRIANGLES
162. Logarithmic Computation. We have already had
occasion to observe that many computations in engineering,
astronomy, etc., are carried out by means of logarithms. Jn
the last chapter a few examples of the use of logarithms in
computation were given in connection with a fourplace table.
Such a table suffices for data and results accurate to four sig
nificant figures. When greater accuracy is desired we use a
five, six, or sevenplace table.
The methods used in connection with such a table differ
slightly from those used ordinarily with a fourplace table.
Accordingly we take up briefly at this point some problems in
volving computation with a fiveplace table of logarithms.
No subject is better adapted to illustrate the use of logarith
mic computation than the solution of triangles, which we shall
consider in some detail. Fiveplace tables and logarithmic
solutions ordinarily are used at the same time, since both tend
toward greater speed and accuracy.
163. Fiveplace Tables of Logarithms and Trigonometric
Functions. The use of a fiveplace table of logarithms differs
from that of a fourplace table in the general use of socalled
" interpolation tables " or " tables of proportional parts," to facil
itate interpolation. Since the use of such tables of proportional
parts is fully explained in every good set of tables, it is unnec
essary to give such an explanation here. It will be assumed
that the student has made himself familiar with their use.*
In the logarithmic solution of a triangle we nearly always
need to find the logarithms of certain trigonometric functions.
* For this chapter, such a fiveplace table should be purchased. See, for ex
ample, The Macmillan Tables, which contain all the tables mentioned here
with an explanation of their use.
IX, § 163] NUMERICAL COMPUTATION 243
For example, if the angles A and B and the side a are given,
we find the side b from the law of sines given in § 125,
, a sin B '
b = — — '
sin A
To use logarithms we should then have to find log a, log (sin B)
and log (sin A). With only a table of natural functions and a
table of logarithms at our disposal, we should have to find first
sin Ay and then log sin A. For example, if A = 36° 20',
we would find sin 36° 20' = 0.59248, and from this would find
log sin 36° 20' = log 0.59248 = 9.77268  10. This double use
of tables has been made unnecessary by the direct tabulation
of the logarithms of the trigonometric functions in terms of
the angles. Such tables are called tables of logarithmic sines,
logarithmic cosines, etc. Their use is explained in any good
set of tables.
The following exercises are for the purpose of familiarizing
the student with the use of such tables.
EXERCISES
1. Find the following logarithms : *
(a) log cos 27° 40'.5. (d) log ctn 86° 53'. 6.
(6) log tan 85° 20'. 2. (e) log cos 87° 6'. 2.
(c) log sin 45° 40'.7. (/ ) log cos 36° 53'.3.
(d) log sin A = 9,78332  10.
(e) log ctn ^A = 0.70352.
(/) log tan J ^ = 9.94365  10.
87325
4. Given a triangle ABC, in which ZA = 32°, ZB = 27°, a  5.2, find
b by use of logarithms.
* Fiveplace logarltbms are properly used when angles are measured to the
nearest tenth of a minute. For accuracy to the nearest second, six places
should be used.
2. Find A
, when
(a) log sin
A = 9.81632 
10.
(6) log cos
A = 9.97970 
10.
(c) log tan ^ = 0.45704.
3. Find^,
iftan. = 4^«\2^
Q7Q.
89.710
244 MATHEMATICAL ANALYSIS [IX, § 164
164. The Logarithmic Solution of Triangles. The effective
use of logarithms in numerical computation depends largely on a
proper arrangement of the work. In order to secure this, the
arrangement should be carefully planned beforehand by con
structing a blank form, which is afterwards filled in. Moreover
a practical computation is not complete until its accuracy has
been checked. The blank form should provide also for a good
check. Most computers find it advantageous to arrange the
work in two columns, the one at the left containing the given
numbers and the computed results, the one on the right contain
ing the logarithms of the numbers each in the same horizontal
line with its number. The work should be so arranged that
every number or logarithm that appears is properly labeled ;
for it often happens that the same number or logarithm is used
several times in the same computation and it should be possible
to locate it at a glance when it is wanted.
The solution of triangles may be conveniently classified
under four cases :
Case I. Given two angles and one side.
Case II. Given two sides and the angle opposite one of the
sides.
Case III. Given two sides and the included angle.
Case IV. Given the three sides.
In each case it is desirable (1) to draw a figure representing
the triangle to be solved with sufficient accuracy to serve as a
rough check on the results ; (2) to write out all the formulas
needed for the solution and the check ; (3) to prepare a blank
form for the logarithmic solution on the basis of these
formulas ; (4) to fill in the blank form and thus to complete
the solution.
We give a sample of a blank form under Case I ; the student
should prepare his own forms for the other cases.
IX, § 165] NUMERICAL COMPUTATION
245
165. Case I. Given two Angles and one Side.
Example. Given : a=430.17, ^=47° 13'.2, B=d2° 29'.6. (Fig. 134)
To find: C, 6, c.
Formulas :
6 =
180°
a
sin J.
a
(A+B),
sin B,
sin C.
Check :
Fia. 134
sin^
cb ^ ts,ni(CB)
c + b tSinl{C\B)'
The following is a convenient blank form for the logarithmic solu
tion. The sign ( + ) indicates that the numbers should be added; the
sign ( — ) indicates that the number should be subtracted from the one just
above it. ^ , ,
Logarithms
A
Numbers
( + )^ =
A + B =
C =
179° 60'
d —
sin^ =
sin
• .
a/sin A
sin B =
b =
sin
• •
•) (+)
■)
a /sin A
sin C = sin
c =.. .
cb =
c+ b =
(>) ( + )
Check
CB=
C + B = . ....
tan^C J5) =tan . . . (
tan^(C5)=tan . . . (
■(1)
(Logs (1) and (2)
. should be equal
.) ( — ) for check.)
(2)
246
MATHEMATICAL ANALYSIS
[IX, § 165
FilUng in this blank form, we obtain the solution as follows.
Numbers Logarithms
A = 47° 13'.2
B= 52° 29\6
A\B= 99°42'.8
179° 60'.0
C = 80° 17'.2
a = 430.17 (>) 2.63364
sin ^ = sin 47° 13' 2 (— >) () 9.8656710
a/sin ^ 2.76797
sin B = sin 52° 29'. 6 (>) ( + ) 9.89943  10
b = 464.94 Ans. (<) 2.66740
a/sin yl 2.76797
sin C = sin80° 17'.2 (>) ( + ) 9.99373
c = 577.70 Ans. (^^) 2.76170
c6= 112.76
c + 6 = 1042.64
(>) 2.05215
(>►) () 3.01818
9.03402  10
CB= 27°47'.6
C + 5 = 132° 46'.8
tan ^ ( a  jB) = tan 13° 53'.8 (>) 9.39342  1
tan 1(0+ B)= tan 66° 23'. 4 (— ^) (_) 0.35942
9.03400  10 J
Check
EXERCISES
Solve and check the following triangles ABC :
1. rt = 372.5, .4 = 25°30', J5 = 47°60'.
2. c = 327.85, vl = 110° 52'.9, B = 40° 31'.7. Ans. C = 28° 35'.4
a = 640.11, b =446.20.
3. a = 53.276, A = 108° 50'.0, C = 57° 13'.2.
4. 6 = 22.766, ^=141°59'.l, a=25°12'.4.
5. 6 = 1000.0, B = 30° 30'. 5, C = 50° 50'.8.
6. a = 257.7, A = 47° 25', B = 32° 26'.
IX, § 166] NUMERICAL COMPUTATION
247
166. Case II. Given two Sides and an Angle opposite
one of them.
If A, a, b are given, B may be determined from the relation
h sin A
(1)
sin B
If log sin B = 0, the triangle is a right triangle. Why ?
If log sin B >0, the triangle is impossible. Why ?
If log sin B <0, there are two possible values Bi, B^ of B,
which are supplementary.
Hence there may be two solutions of the triangle. (See Ex. 1,
page 249.)
No confusion need arise from the various possibilities if the
corresponding figure is constructed and kept in mind.
It is desirable to go through the computation for log sin B
before making out the rest of the blank form, unless the data
obviously show what the conditions of the problem actually
are.
Example 1. Given : A = 46° 22'.2, a = 1.4063, 5=2.1048. (Fig. 135)
To find: S, O, c.
Formula : sin B
b sin A
a
Fig. 135
Numbers
& = 2.1048 (
sin ^ = sin 46° 22'. 2 (
bsinA
a = 1.4063 (
sin B {<
Logarithms
0.32321
►) ( + ) 9.85962 ~ 10
0.18283
() 0.14808
.) 0.03475
Hence the triangle is impossible. Why ?
248
MATHEMATICAL ANALYSIS
[IX. § 166
Example 2.
Given : a = 73.221, b = 101.53, A ■.
To find: J5, C, c.
40° 22'.3. (Fig. 136)
Formula: sin 5 =
6sin^
Numbers
b = 101.63 (
sin^=sin40°22'.3 (
b sin A
a = 73.221
sin JB
G
Logarithms
2.00660
) ( + ) 9.81140 10
11.81800  10
1.86464
(^) ()
9.95336  10
The triangle is therefore possible and
has two solutions (as the figure shows).
We then proceed with the solution as
follows :
We find one value Bi of B from
the value of log sin B. The other
L JJ lO llUCJU j
Other formulas :
(7=180°(^ + B). .
asin C
sin^
Check: ^^:
c + &
tanUCB)
tanKO + 5)
Numbers
Logarithms
sin B
9.95336  10
Bi= 63°65'.2
179*^ 60'.0
^2 = 116° 4'.8
4 + 5i = 104° 17'.6
179° 60'.0
Ci= 76'^42'.6
a
(»
1.86464
a'mA
(» (
) 9.81140
a/sin A
. 2.06324
sinCir=8in76°42.'6
'(>) ( + ) 9.98634]
Cl
10
10
109.64 (^) 2.03968
IX, § 167] NUMERICAL COMPUTATION
249
Ci  6 = 8.01 (.
Ci + 6 = 211.07 (.
Ci^i= 11°47'.8
Ci + ^i = 139°37'.7
tan K Ci  Bi) = tan 6° 63'.6
tan K Ci + Bi) = tan 69=^ 48'.8
0.90363
.) () 2.82443
8.5792010
9.01377  10
0.43455
8.57922  10
Check*
One solution of the triangle gives, therefore, B=63° 55'.2, C = 76° 42'.5,
c = 109.54.
To obtain the second solution, we begin with B2 = 116° 4'. 8. We find
C2 from C2 = 180° (A + B^) ; i.e. C% = 23° 32'. 9. The rest of the com
putation is similar to that above and is left as an exercise.
EXERCISES
1. Show that, given A, o, 6, if A is obtuse, or if A is acute and a > 6,
there cannot be more than one solution.
Solve the following triangles and check the solutions :
2. a = 32.479, 6 = 40.176, ^ = 37° 25M.
3. 6 = 4168.2, c = 3179.8, B = 51° 21'.4.
4. a = 2.4621, b = 4.1347, B = 101° 37'.3.
5. a = 421.6, c = 532.7, ^ = 49° 21'.8.
6. a = 461.5, c= 121.2, C = 22° 31'.6.
7. Find the areas of the triangles in Exs. 26.
167. Case III. Given two Sides and the Included Angle.
Example. Given: a=214.17, 6=356.21,
C=62°21'.4. (Fig. 137)
To find : A, B, c.
Formulas :
6
tan I {B A)
tan KB + A);
b + a
B+A= 180°  C = 117° 38'.6 ;
_ gsin C _ 6 sin C
sin A sin B
* A small discrepancy in the last figure need npt cause concern. Why?
250
MATHEMATICAL ANALYSIS
[IX, § 167
Numbers
6  a = 142.04 (
6 + a = 570.38 (
(b  a)/{h + a)
tan ^{B + A) = tan 58° 49' .3 (
tan \{B A) = tan 22° 22^2 (
.•.A= 36°27M
5= 81°11'.5
)
)
Ans.
Ans.
a = 214.17 (>)
sin ^ = sin 36° 27'. 1 (>)
a/sin ^
sin C = sin 62° 21 '.4 (>)
c = 319.32 Ans. (^
Check by finding log (6/sin B).
Logarithms
2.16241
() 2.75616
9.39625  10
( + ) 0.21817
9.6144210
2.33076
() 9.77389 10
2.55687
( + ) 9.94736 10
2.50423
EXERCISES
Solve and check each of the following triangles.
1. a = 74.801, h = 37.502, C = 63° 35'.5.
2. a = 423.84, 6 = 350.11, C = 43°14'.7.
3. 6 = 275, c = 315, ^ = 30° 30'.
4. « = 150.17, c = 251.09, ^ = 40°40'.2.
6. a = 0.25089, b = 0.30007, C = 42° 30' 20".
6. Find the areas of the triangles in Exs. 15.
168. Case IV. Given the three
Sides.
Example. Given : a = 261.62,
b = 322.42,
c = 291.48.
To find : A, B, G.
Formulas :
s = l{a \ b \ c).
yJ (« — q)(gft)(gc)
tan ^A= — ^ , tan I 5 = **
a s — b
Check: A \ B \ C = 180°.
tan I C =
b=3gg.4g A
Fig. 138
8 — C
(§ 143)
IX, § 168] NUMERICAL COMPUTATION 251
Numbers
a = 261.62
6 = 322.42
c = 291.48
( + )
()
Logarithms
2s = 875.52
s = 437.76
sa= 176.14 (>)
sb= 115.34 (^)
sc = 146.28 (^)
2.24586
2.06i98
2.16518
2s = 875.52 (Check.)
8 = 437.76 (>)
6.47302
2.64124
r
s— a
3.83178
1.91589
2.24586
tan ^ ^ = tan 25° 4'.1 (<)
9.67003  10
r
sb
1.91589
2.06198
tan 1 5 = tan 35° 32'.4 (<— )
r =
s— c =
9.8539110
1.91589
2.16518
tan ^ C = tan 29° 23'. 4+ (<)
A = 50° 8'. 2 ^ws.
B= 71° 4'.8 ^ns.
0= 68° 46'. 9 Ans.
9.7507110
179° 59'. 9 Check.
EXERCISES
Solve and check each of the following triangles :
1. a = 2.4169, b = 3.2417, c = 4.6293.
2. a = 21.637, & = 10.429, c = 14.221.
3. a = 528.62, b = 499.82, c = 321.77.
4. a = 2179.1, & = 3467.0, c = 5061.8.
5. a = 0.1214, 6=0.0961, c = 0.1573.
6. Find the areas of the triangles in Exs. 15.
7. Find the areas of the inscribed circles of the triangles in Ex. 16.
252
MATHEMATICAL ANALYSIS [EX, § 169
III. THE LOGARITHMIC SCALE — THE SLIDE RULE
169. The Logarithmic Scale. Let us lay off, on a straight
line, segments issuing from the same origin and proportional
to the logarithms of the numbers 1, 2, 3, 4, —. The base of
the system of logarithms is immaterial. Let us label the end
points of these segments by the corresponding numbers. This
gives a nonuniform scale, which is called a logarithmic scale.
Such a scale is pictured in Fig. 139.
[
T
s 4
Fig. 139
rm
A scale of this kind is easily constructed from the graph of
the logarithmic function (Eig. 133).
170. The Slide Rule. The slide rule is an instrument often
used by engineers and others who do much computing.* It
consists of a rule (usually made of wood faced with celluloid)
Fig. 140
along the center of which a slip of the same material slides
in a groove. This slip is called the slide. The face of the
slide is level with the face of the rule.
* Engineers usually purchase rather expensive slide rules made of wood
and celluloid. These are on sale in all stores which carry draftsmen's supplies.
A very simple slide rule sufficiently accurate for class purposes is printed on
hard pasteboard and is obtainable at reasonably small cost through any one
of several manufacturers of instruments. Figure 140 is reproduced on a larger
scale on the first flyleaf at the back of the book. By cutting out this leaf
and carefully cutting up the figure, a slide rule can be made by the student.
This will not be very accurate, but it will suffice to illustrate the principles.
XI, § 170] NUMERICAL COMPUTATION
253
Along the upper edge of the groove are engraved two loga
rithmic scales, usually labeled A and B, the scale A being on
the rule, the scale B on the slide. (See Fig. 141.)
The scales A and B are identical. The slide is simply a
mechanical device for adding graphically the. segments on
r
1 2
C
I 5 6 7 8 9 1
^ 1 I 1 1 I 1 , , , , 1
2
^ ihiililijlilililiilililili
[llllili
m
m
m
w'm
Tn:
■]
trhl
1
\ f v>
1 1
1
m
lU
ill
1 p
M 1 1
:::nn]iiioi
1
\
C
N
tn , ,,,,, ...^Tg; ■ 4 ■ ,
) 6
7 8 9
I
2
3
y
iTi"TiiiTr
'IHW
]]Xii"":i
"■""I""I]
^iiiiiiiiiiiiiiiiiiiiiiiiiii
U.iuM.ljLi
' 1
..Mr.i:ii
:ii:i4
1 2
J
4
Fig. 141
these scales. Since the segments represent the logarithms of
the numbers found on the scale, the operation of adding the
segments is equivalent to multiplying the corresponding num
bers. Thus, to find the product 2.5 x 3.2 move the slide to the
right until the point marked 1 at the extreme left of the
slide (scale B) is in contact with the point 2.5 on scale A
(Fig. 141 shows the positions of scales A and B after this
operation). The point 3.2 on scale B is then opposite the point
8.0 on scale A. The latter number is the required product :
2.5 X 3.2 = 8.0. A little reflection should make quite clear
how the operation just performed is equivalent to adding the
logarithms of 2.5 and 3.2 and then reading from the scale the
number corresponding to the sum. We may note further that
with slide set as in the example just worked it is set for
multiplying any number by 2.5 ; i.e. every number of the scale
A is the product of 2.5 by the number below it on scale B.
The slide is therefore also set for division by 2.5. Every
254 MATHEMATICAL ANALYSIS [IX, § 170
number of scale B is the result of dividing the number above
it by 2.5. Thus we read from the scale (set as before) that
7.2 ^ 2.5 = 2.9 approximately.
Having now shown very briefly how the slide rule may be
used for multiplication and division, let us examine it a little
more closely. Scales A and B are labeled with the numbers
1,2, 3, 4, 5, 6, 7, 8, 9, 1,2, ...,9,1.
It is natural to ask why the number following the 9 in the
middle of these scales is not labeled 10 ? The answer is that
the numbers on the slide rule are given without any reference
to the position of the decimal point, just as the numbers in a
table of logarithms are given without reference to the decimal
point. The number 1 at the extreme left of the scale may
represent either 1, or 10, or 100, or 1000, etc., or .1, or .01, or
.001, etc. If the 1 at the extreme left of the scale represents
1, then the other numbers on the first half of the scale repre
sent 2, 3, ..., 9, the 1 in the middle represents 10, the 2 represents
20, and the successive numbers represent 30, 40, ., 100 (the
last being represented by the 1 at the extreme right of the
scale). If on the other hand the 1 at the left represents 100,
the successive numbers represent 200, 300, ..., 900, 1000, 2000,
..., 10,000. If the 1 at the left represents .1, the successive
numbers represent .2, .3, .••, .9, 1.0, 2.0, ••., 10.0 ; and so on.
The reading of the subdivisions on the scales (A and B)
should now offer little difficulty. Whenever an interval be
tween two successive numbers is divided by certain lines of the
same length into 10 parts, each of these parts represents one
tenth of the number represented by the interval in question.
Thus, if we fix our attention on the division between 2 and 3,
we note that a certain set of lines divides this interval into 10
parts ; if the 2 represent 2, these divisions represent respec
tively 2.1,2.2, ..., 2.9. On the other hand, if the 2 is thought
IX, § 170] NUMERICAL COMPUTATION 255
of as representing 20, these divisions represent 21, 22, —, 29 :
and so on. These divisions into ten are at some parts of the
scale subdivided further into five or two parts. These parts
then represent fifths or halves of the interval that represented
a tenth. Thus we may readily locate on the scale the point
representing 1.42 or the point representing 3.65.
Turning our attention to scales C and D along the lower
edge of the groove on the slide and the rule respectively,
we note first that these two scales are also identical. Compar
ing them with scales A and B, we see that the unit chosen for
C and D is just twice the unit of A and B. Hence the scales
C and D can be used for multiplying and dividing just as
scales A and B are used ; however on C and D our range is
smaller. The range of numbers on A and B is from 1 to 100 ;
on C and D only from 1 to 10. To make up for this limitation,
scales C and D give greater accuracy.
However, the principal reason for the existence of the second
pair of scales is the fact that the two pairs of scales thus ob
tained furnish a table of squares and square roots. In view of
the relation between the units with respect to which the two
pairs of scales are constructed, every number of scale A is the
square of the number vertically below it on scale D. Why ?
In order that corresponding numbers on scales A and D may be
accurately read off, every slide rule is provided with a runner,
the vertical line on which connects corresponding numbers of
the upper and lower scales. The runner also enables us to
perform calculations consisting of several operations without
reading off the intermediate results, thus saving time and
securing greater accuracy in the final result. The actual use
of the slide rule will be explained in the next article.
The successful use of the slide rule depends largely on the
ability to read the scales readily and accurately, accuracy
256 MATHEMATICAL ANALYSIS [IX, §170
often necessitating the estimating of numbers falling between
the lines of division. The ability mentioned can be secured
only by practice. A proficient operator, with a teninch slide
rule, can always secure results accurate to three significant
figures. This degree of accuracy is sufScient for many of the
computations of applied science, manufacturing, etc., in which
the slide rule is proving more and more useful.
171. The Use of the Slide Rule. All calculations in mul
tiplication, division, proportion, etc., are worked on scales Cand
D unless the answer is so large that it does not lie on the scale.
In that case scales A and B are used. Let us begin with pro
portion. On this topic, and on the corresponding property of
the slide rule, all computations involving multiplication or
division, or both, maybe made to depend in a very simple way.
The property of the slide rule referred to is as follows : No
matter where the slide be placed, all the numbers on the slide
bear the same ratio to the corresjjonding numbers on the rule (due
regard being had to the position of the decimal point). For
example, if the slide be set so that 2 of O coincides with 4 of
D, it will be observed that the same ratio 2 : 4 exists between
every pair of corresponding numbers : 1 : 2, 3 : 6, 42 : 84,
125 : 250, etc. Explain why this is true. This leads at once to
the rule for finding the fourth term of a proportion, when the
first three are given. We give this rule in diagrammatic form,
as follows : *
To find the fourth term of a proportion :
c
D
Set first term
over second term.
Under third term
find fourth term.
♦ Iji this article we have followed to a considerable extent the treatment
given in the Manual for the use of the Mannheim Slide Rule, published by
the Keuffel and Esser Co., New York.
IX, § 171] NUMERICAL COMPUTATION
This gives the solution of the equation
b X
To find the product abj solve the proportion
a X
To find the quotient , solve the proportion
a__x
The following examples will make clear the procedure.
Example 1. Solve the proportion : 13/24 = 32/a;.
257
D
Set 13
over 24
Under 32
find 69.1 Ans.
Example 2. Solve the proportion : 13/24 = 75/x.
Since the first two terms of the proportion are the same as in the pre
ceding example, we set the slide as before. We now find, however, that
75 on C is beyond the extremity of D. We accordingly set the runner on
the lefthand 1 of 0, and then set the righthand 1 of C on the runner.
We find under 75 the number 138.5, the required value of x* (Justify
the above use of the runner. )
The same example can be done on scales A and B with one setting,
without using the runner.
Example 3. Find the product: 23.2 x 5.3.
c
D
Setl
over 23.2
Under 5.3
find 123.0 Ans.
Here we set the righthand 1 on 23.2. Use whichever 1 serves. The
decimal point, in this as in the other examples, is simply located by in
spection and a brief mental estimate of the answer. Here we see readily
that the answer is something over 100; hence we locate the decimal
point at the place to give us 123.0.
* The .5 in this answer must be estimated. Usually, if more than three
significant figures are obtained from the rule, the last is uncertain.
258 MATHEMATICAL ANALYSIS [IX, § 171
Example 4. Find the value o/364 4 115.
c
D
Set 364
over 115
Find 3.17, Ans.
over 1
Example 5. Find the circumference of a circle whose diameter is 42
ft. We multiply the diameter by tt = 3.14.* Hence,
c
D
Set 1
over 3.14
Under 42
find 132.0 Ans.
By ordinary multiplication we get 131.88 ; an example of the inaccur
acy of the fourth significant figure.
Example 6. Find the continued product : 1.6 x 4.2 x 5.3 x 2,8.
The abbreviation R. denotes the runner on the sliderule.
Set 1
over 1.6
R, to 4.2
1 to R.
R. to 5.3
1 toR.
Under 2.8
find 99.7 Ans.
We add a fevnr more rules for computing various types of expressions
involving scales A and B as vv^ell as C and D.
(1) To find a^ xb:
(2) To,
A
Find a2&.
Ans.
B
over b.
C
Setl
D
over a
ada'^h
b:
A
Find a2 h b,
Ans
B
Set 6
over 1.
C
D
over a
* The number t is usually raa>'ked on the scale.
IX, § 171] NUMERICAL COMPUTATION
259
(3) To find geometric mean between two numbers a and b; i.e. lincl x,
so that a/x = x/b. Let a < 6.
A
B
Seta
Below 6
C
D
over a
findiX=G.M:
(4) To reduce fractions to decimals
Set numerator
over denominator
Find equivalent decimal
above 1
These rules are not to be memorized. They will be used almost in
stinctively by one who has made the reason for each rule thoroughly clear
to himself and who is in practice.
EXERCISES
1. With a slide rule compute the value of :
(a) 2.13 X 4.42. {h) 2,856,000 x 256,700,000.
(&) 1.98x5.24., ___ 5,43^31.5
(c) 2.77 x 3.14 X 4.25.
id) 8.27/2.63.
(e) 5.48/3.26.
(/) 10/3.14.
{g) 0.000116 X 0.0392.
(0
U)
21.4
7.64 X 4.14
21.2
67.4 X 25.5 X 19.7
4.64 X 18.4
2. With a slide rule compute the value of :
(a) (2.85)2. (c) (1.86)3.
(6) 3.72 X (2.23)2. (^) (6.24)2/26.3.
3. Find the circumference and the area of a circle whose radius is
4.16 in.
4. What is the length in feet of 27.3 meters, given that 26 meters =
82 feet ? Solve with one setting of the slide.
260 MATHEMATICAL ANALYSIS [IX. § 172
IV. LOGARITHMIC PAPER
172. Logarithmic Paper. Euled paper is printed, on which
the rulings in both directions are spaced according to the
logarithmic scale (§ 169), i.e. precisely as on a slide rule.*
Such paper is called logarithmic paper. Samples of this ruling
are shown in Figs. 142143.
173. Plotting Powers on Logarithmic Paper. The graphs
of equations of the type
(1) y = kx""
can be plotted very readily on logarithmic paper. For, if we
take the logarithms of both sides, we find
(2) log 2/ = log A: h n log x.
Let us set Y=\ogy, K=\ogk, X=loga;;
then (2) becomes
(3) Y=K\nX.
Now the equation (3) represents a straight line if X and Y be
taken as the variables. This is precisely what happens if we
plot the values of x and y from equation (1) on logarithmic
paper ; for, when we plot a value for x on logarithmic paper, the
distance from the left border is nothing else than logic, i.e. X;
and similarly for Y.
Moreover, the slope of the straight line represented by (3) is
n, the exponent of x in (1) ; and the intercept on the Y axis is
K= log k. Hence if values of x and y from (1) are plotted on
logarithmic paper, the value of n in (1) appears as the slope of
the straight line graph, and the value of k can be read off
directly on the vertical axis.
* On this account, it is possible to make a crude slide rule by using the
edges of two sheets of logarithmic paper, sliding them along each other after
the manner of a slide rule.
IX, § 173] NUMERICAL COMPUTATION
261
Example 1. Draw the graph of the equation y = r^ on logarithmic
paper.
Take x — \, then y =\. Take x = 10, then y = 100. Plot these two
points A (1, 1) and B (10, 100) (Fig. 142). Connect A and 5 by a
straight line. This is the required graph.
The graph may be drawn also by noticing that its slope is the exponent
ifm
11,1111 1 1 1 1 1 1 11
y!:::::::::: :
o
II II 1
t
2 ___
Q
5 
::::::: :: t
/
::::::: :: i
f
3 
7
t
....
7
::::::: ^
1.5
2 L
::::::: ^E= —
::::::: izzz zi
7
9
17
z
^
r^
::::::7 ::
/
/
J
f
__
t
'i
_
La?
Al
1.5 2
4 5
7 S91
1.5
4 5 6 7 891
Fig. 142
of X in the given equation, i.e. 2. Hence we may draw from A a line
whose slope is 2. Show that this gives the same line, AB.
We may use this graph to find squares or square roots. Thus, if a; = 4,
we can note the point on the graph directly over 4, and read the_corre
sponding value of y, which is 16. Reversal of the process gives \/16 = 4.
Likewise, if x = 4.5, we find y  20.2+ ; and vl5 = 3.8, approximately.
262
MATHEMATICAL ANALYSIS
[IX, § 173
Conversely, given a straight line on logarithmic paper, we
know that its equation must be of the form (1). We can j&nd
n by actually measuring the slope, and we can read off k on the
vertical line through the point marked (1, 1), since if we place
lOjYTTl
—
—
7
9 —
t
n
7 "
/
«
^
/
J.
/
■y
" e= elongation in c
i^r^pull in kg.
y
m.
/
i
2
' e = .3
r
J
f
1.5
^
^
' —
—<
«J —
/
/
_ TZ
r
■y
c
/
/
/
I
~~T
k^S,/__
1
.2
=
=
.15
E
i^A
E
1
■
1
17
X
.iLLLLL
_J
1
,15 .2
.3
.4 .5 .6 .7.8.91 1.5
Fig. 143
3 4 5 6 7 8 910
a; = 1 in equation (2), we have log?/ = logfc, whence 2/ = A;.
Any other value of x may be used instead of £c = 1, but a; = 1 is
most convenient because log 1 = 0.
Example 2. A strong rubber band stretched under a pull of p kg.
shows an elongation of e cm. The following values were found in an
experiment :
IX, § 173] NUMERICAL COMPUTATION
263
p
0.5
i:o
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
6.0
7.0
e
0.1
0.3
0.6
0.9
1.3
1.7
2.2
2.7
3.3
3.9
5.3
6.9
If these values are plotted on logarithmic paper, . it is evident
that they lie reasonably near a straight line, such as that drawn in
Fig. 143.
By measurement in the figure, the slope of this line is found to be 1.6,
approximately. Hence if we set
P= logp, ^ = 'loge,
we have ^=^+1.6P,
where iTls a constant not yet determined ; whence
loge = K{ \.Q\ogp
or . e = kp^^^
where K = log k. If p = 1, e = A; ; from the figure, if j> = 1, e = 0.3 ;
hence k = 0.3, and
e = 0.3pi6.
EXERCISES
1. Plot on logarithmic paper the graph of each of the following equa
tions :
(a) y — 7?. (c) y = afi. (e) y = S x^.
(b) y = xi (d) y = x^s. (/) y = 4.5 x^^.
2. Draw the graph ot y = x^. Note that the negative exponent — 2
gives simply what we ordinarily call a negative slope of — 2 for the
straight line graph.
3. When air expands or is compressed (as in an air compressor) , with
out appreciable loss or gain of heat, the pressure p and the volume v are
connected by the formula
p = kv~^^, approximately.
Pressure is often measured in atmospheres, and volume in cubic feet.
If we start with one cubic foot of air at one atmosphere of pressure, it is
obvious that k = 1. Draw the graph for this case, and from it find p
when V = 0.5 cu. ft. Find v when p = 5 atmospheres. Find v when
p = 0.5 atmospheres.
264
MATHEMATICAL ANALYSIS [IX, § 173
4. The intercollegiate track records for foot races (1916) are as
follows, where d means the distance run, and t means the record time :
cl
100 yd.
220 yd.
440 yd.
880 yd.
Imi.
2 mi.
t
0:09
0:21^
0:48
l:54f
4 : lof
9:23f
Plot the logarithms of these values on squared paper (or plot the given
values themselves on logarithmic paper). Find a relation of the form
t = k(P^. What should be the record time for a race of 1320 yd.?
(See Kexnelly, Popular Science Monthly, Nov. 1908.)
5. In each of the following tables, the quantities are the results of
actual experiments ; the two variables are supposed theoretically to be
connected by an equation of the form y = fccC*. Draw a logarithmic graph
and determine k and ?i, approximately :
(a) (Steam pressure ; v = volume, p = pressure. )
V
2
4
6
8
10
p
68.7
31.3
19.8
14.3
11.3
(Saxelby.)
(6) (Gas engine mixture ; notation as above.)
V
3.54
4.13
4.73
5.35
5.94
6.55
7.14
7.73
8.05
p
141.3
115
95
81.4
71.2
63.6
54.6
50.7
45
(Gibson.)
(c) (Head of water h, and time t of discharge of a given amount.)
h
0.043
0.057
0.077
0.095
0.100
t
1260
540
275
170
138
(Gibson.)
CHAPTER X
THE IMPLICIT QUADRATIC FUNCTIONS
Twovalued Functions
I. THE FORMS Ax^^ Ey+ C = AND By'^ { Dx { C =
174. The General Implicit Quadratic Function. We shall
now return to the discussion of algebraic functions. We first
discussed the explicit linear function y=mx\b, and the function
y defined by the implicit relation Ax f By +0 = (Chapter
III). Then we discussed the explicit quadratic function of
the form y = ax"^ \hx \ c (Chapter IV). We now propose to
take up the discussion of the functions y defined by implicit
quadratic relations, such as 4 !/2 — 5 a? = 0, ic^ — 4 ?/f 2 a;— 41/— 1
= 0, etc. The most general form of such an equation is
(1) Ax"^ + Fxy + By^ + Dx } Ey + C = 0.
The graphs of equations of this form are important curves,
with interesting geometric properties, which we shall discuss
in a later chapter. Our present purpose is to determine the
general nature of these graphs (their shape, etc.) and to develop
methods whereby the graph of a given equation of the type
considered may be readily drawn.
We may note at the outset that the function defined by an
implicit quadratic relation between x and y will usually be
twovalued, i.e. to each value of x will correspond, in general,
two distinct values of y. This is due to the fact that if any
particular value be assigned to x in equation (1) above, the
265
266 MATHEMATICAL ANALYSIS [X, § 174
corresponding values of y are determined by a quadratic equa
tion, unless ^ = 0.
We shall approach the discussion of equations of type (1) by
considering in order certain simpler forms of this general
type. First, we shall discuss equations of the two types
Ax'+Ey +C = and By'^ ^ Dx \ C =^ 0.
175. The Equations x'^ — y — and y'^—x—0. We can dis
pose of the equations x"^ — y = and 3/2 — a; = very quickly.
The first equation is equivalent to the equation y = x^, already
discussed in § 72. The second equation is equivalent to the
equation
(2) y^=x,
or y = ± Va;.
We can either plot the points {x, y) whose coordinates satisfy
this relation and thus obtain the graph desired * ; or, we can
note that the equation y"^ = x is obtained from the equation
352 = ?/ by simply interchanging x and y. Hence, the graph of
y"^ = X is obtained from the graph of y = x^ by turning the
plane of the graph oi y — x"^ over about the line through the
origin bisecting the first and third quadrants. Eor, this opera
tion will interchange the x and ?/axes in the desired way. The
two graphs are shown in Fig. 144.
Certain properties of the graph of the equation y"^ = x are at
once evident from the form of the equation : The graph is
symmetric with respect to the a^axis ; for, if a point (Ji, k)
satisfies the equation, the point {Ji, — k) also satisfies the
equation. Why ? The graph lies at the right of the ajaxis ;
for, any negative value of x would give rise to imaginary
values, of y. Why?
 * A table of square roots will facilitate the work.
X, § 176] IMPLICIT QUADRATIC FUNCTIONS
267
I
i
i
t ,i / 3
:_:_V — ^ :: = =— 
5 , S 1 ^'^^"'
\ 7A^ \
^/1v_ __ __ _._ _.
  1^ ix^ ::fe^ . ir 2' ■■ ^"^^L
^s
4^5; " ~ : J".^ :
r ._ ..,,, ^f^ "^ . .
__±i_J 1 1 1 1 M 1 1 M 1 i 1 11 1 1 1 M 1 m 1 M
Fig. 144
The most important properties of the doublevalued function
± Va; to be noted are the following :
(1) For every positive value of x there are two values of the
function, viz. f V^ and — ^x. Therefore the function is two
valued.
(2) As X increases numerically, the corresponding values of
■\Jx increase numerically, i.e. the numerical value of Va; is an
increasing function of x.
176. The Form By'' f Dx = 0. B^^.
may always write the equation in the form
(3)
Since B^^^ we
B '
Y
^

n>o
X
y'
i.e. in the form
where n = — D/B. The graph is
then similar to that of x^^ny, the
only difference being that the roles
of the X and i/axes are interchanged. If the coefficient n is
positive, the graph is at the right of the 2/axis ; if n is nega
tive, the graph is at the left of the .vaxis (Fig. 145). In both
cases the graph is symmetric with respect to the icaxis, and
Fig. 145
268
MATHEMATICAL ANALYSIS
[X, § 176
passes through the origin, at which point it has a vertical tan
gent. Why ? The curve defined by an equation of the type
considered is called a parabola if D =^0. (See Chapter IV.)
To sketch such a curve rapidly, knowing its general shape, we
need only plot a few corresponding values of x and y. If i>=0,
the equation becomes By'^=0. Its graph is then the a^axis.
177. The Slope of the Curve By^ + Dx = 0. To determine
the slope of the tangent to the curve
By^{ I)x= 0,
we may proceed by the method used for similar problems in
Chapters IV and V. To this end we first calculate the change
ratio Ay /Ax, which is the slope of the chord PQ (Fig. 146). The
z^
\^Au
TTo
Fig. 146
slope of the tangent at P is then the limit which this ratio
approaches when Ax approaches the value 0.
Let P(£Ci, 2/i) be any point on the curve, and Q(a;i + Ax, t/i + Ay)
be another such point. Then we have
J5(2/i + AyY + i)(a!i + Ax) = 0,
and
Expanding the first of these equations, and subtracting the
second from it, we get
2 By Ay + BAy'^ + DAx = 0,
or
{2By,rBAy)^^=D.
X, § 177] IMPLICIT QUADRATIC FUNCTIONS 269
Hence, the desired change ratio is
Aj/_ D
Aa? 2Byi\BAy°
When Ax approaches zero, Ay also approaches zero. Why?
The desired slope of the curve
Bi/ + Dx =
at the point (xi, y^ is, therefore,
^^ 2By,
The expression for the slope exhibits certain properties of
the curve :
(1) The curve has a vertical tangent at the origin (2/1 = 0).
(2) The slope of the curve above the it'axis is positive, if B
and D have opposite signs ; and negative, if B and D have the
same sign.
(3) The slope of the curve decreases indefinitely in absolute
value as the point (a^i, 2/1) recedes indefinitely from the origin.
EXERCISES
1. For each of the following equations, determine the slope at the point
(xi, y\) and sketch the curve represented. For each point plotted deter
mine the slope of the tangent and draw the tangent.
(a) 2/24x^:0; (6) ?/2 + 2a; = 0; (c)4x23y = 0;
((?) 4 2/2 + 9x^0; (e) y^ = Qx.
2. Derive the equation of the tangent to each of the curves in Ex. 1 at
the point indicated :
(a)(l,2); (6)(_2,2); (c)(3,12); (d) (4,  3) ; (e)(6,6)'.
3. Show that the equation of the tangent to the curve y2 = 2 px at the
point (xi, yi) on the curve is y\y =p (x + xx).
4. Draw the curves y'^ = nx for several different values of n on the
same sheet of paper. It is suggested that the values w = l,2, 5, — 1, — 2,
be included.
270 MATHEMATICAL ANALYSIS pC, § 178
II. THE FORM Ax"^ + By^ + C=
178. The Case A=^B. The Equation x"" ^ y^ = a\ It so
happens that, if the units on the x and ?/axes are equal, we can
interpret the lefthand member of this equation geometrically.
For, it is evident from the figure (Fig. 147) that, under the
Fig. 147
hypothesis of equal units, x^ \ y^ is the square of the distance
of the point (x, y) from the origin. Hence the equation
(5) a;2^2/' = «^
states that the point {x, y) is distant a units from the origin.
It follows that the points {x, y) satisfying this equation are all
on the circle described about as center with the radius a, and
conversely the coordinates of every point on this circle will
satisfy the equation. The graph of the equation x^ \y'^= a^ is
then a circle, if the units on the two axes are equal.
If the units on the axes are unequal, the ordinates of the
above circle must be shortened or lengthened in a certain ratio,
according as the unit on the ^/axis is less than or greater than
the unit on the a^axis. In either case the graph of the equa
tion will be a closed curve.
Throughout the remainder of this chapter, however, we shall
assume, in order to fix ideas, that the units on the axes are equal.
If A = B {AB ^ 0), the equation
(6) A^^Bf^Q^^
X, § 179] IMPLICIT QUADRATIC FUNCTIONS
271
may be written in the form x^ \ y^ = — — »
The graph of this equation is a circle, if — C/A is positive. If
— C/A is negative, the equation has no graph, i.e. no pair of
real values of x and y can satisfy it. If C — 0, tHe only point
satisfying the equation is the origin.*
179. The Case A >0, B >0. Consider first the special
case x"^ + 4t y = 9. If we solve this equation for ?/, we have
(7)
y = ±^^9x\
Now, we know from § 178 that the graph of the function
(8) 2/ = ±V9^=^
is a circle with center at the origin and radius equal to 3.
Y
^^ '~~^
^ \
>^ "^i
' ~ Oliji X
y> VV
T "^^ ^^ Z
. ,
X .^
■^^ ^^
Fig. 148
The ordinates of the points of (7) are then equal to one half
the corresponding ordinates of the points on the circle (8). The
construction of the graph of (7) should then be clear from the
figure (Fig. 148). The graph in question is a closed curve,
having a greatest length of 6 units and a greatest width of 3
units. It is symmetric with respect to both axes.
* The last locus may be considered as a circle with radius equal to 0; it is
sometimes called a poini circle.
272
MATHEMATICAL ANALYSIS
[X, § 179
(9)
The general form
Ax^^ By^jC=0
can be treated similarly, if A and B are both positive,
equation may be written in the form
C
A
The
(10)
x^ + ?f =
This shows that there is no graph if the righthand member is
negative. If the righthand member is 0, the point (0, 0) is the
only point satisfying the equation. There remains only the
case where — C/A is positive.
Equation (10) gives _
(11) ,=±^.^__.,
Now, the equation
(12)
y
represents a circle. Equation (11) tells us that the desired
graph is obtained by shortening or lengthening the ordinates
of this circle in the ratio ^A/B to 1.
I >^fTTNL I I I I I I Example. If we solve the equation 9x^
7^^1l^^tIIII \4y^ = S6 for?/, we obtain y = ± V4 — x^;
/[ I mI I I r\ I I I I this tells us that the graph of the given
equation is obtained from that of the circle
y = ± V4 — x'^ by lengthening the ordinates
of the latter to three halves their original
length. Figure 149 exhibits the result.
mm
The graph of an equation of the form
Fig. 149 Ax^\By^}C=0 under the hypothesis
that A and B are both positive and that C is negative, is then
a closed curve symmetric with respect to both axes.
The curve represented by an equation of the form (9) above
is called an ellipse. An ellipse is symmetric with respect to
X, § 180] IMPLICIT QUADRATIC FUNCTIONS 273
each of two perpendicular lines, called the axes of the ellipse.
The intersection of the axes of an ellipse is called the center
of the ellipse. Knowing the general shape of the curve, the
quickest way to sketch it from the equation is to find the
intercepts on the axes and draw a symmetric curve through
the four points thus obtained. In the example 9 x^ \ 4: y"^ == 36
already considered, we find the intercepts to be ic = ± 2 (found
by placing 2/ = 0) and y = ±o (when x = 0). If we mark the
four corresponding points, the curve can be sketched readil}.
EXERCISES
1. Discuss the locus of each of the following equations and, if the
equation has a locus, sketch it and show how it is related to a certain
circle (if the locus is not itself a circle) :
(a) x^ \ y^ = 16. (d) ix^\y^\16 = 0. (g) ix^ \ Sy^ = 12.
(c) 4 0:2 42/216=0. (/) 2x^{2y2 = 5. ^ ^ 4 "^ 9
2. For what values of x in each of the equations in Ex. 1 doesy become
imaginary ? For what values of y does x become imaginary ?
3. Show directly from the equations that each of the graphs in Ex. 1,
if it exists, is symmetric with respect to both the xaxis and the yaxis.
4. According to the definition above, is a circle an ellipse ?
180. The Slope of the Curve represented by Ax^ + By^
\ C = 0, Here again we calculate the change ratio Ay /Ax,
which is the slope of the secant joining the points P(xi, y^) and
Q{xi + Aa;, y^ + Ay) on the curve, and then find the limit which
this ratio approaches when Q approaches P along the curve, i.e.
when Ax and, consequently. Ay approach 0. The calculation is
as follows :
Since P and Q both lie on the curve
, Ax' + By^+C^O,
we have
(13) Ax^^{By,^+C=0,
T
274 MATHEMATICAL ANALYSIS [X, § 180
and
(14) A{x, + Axy + B(y, + Ayf +C=0.
Expanding the squares in the last equation and subtracting
(13) from (14), we have
2 Ax^Ax + AAx"^ + 2 By^Ay + BAy^ = 0,
or (2 By^ + BAy) Ay = — {2 Ax^ + AAx) Ax,
whence we obtain the slope of the line PQ,
Ay _ _ 2 Axi f AAx
Ax~ 2 By I + BAy
{B=^0).
When Ax and Ay both approach 0, we get for the slope of the
curve at the point (xi, y^)
(15) m = ^.
«
An interesting verification of this result may be noticed. It is well
known that the tangent to a circle at a point P is perpendicular to the
radius OP. Now consider a circle with center at the origin. The slope of
the radius through (xi, yi) is then clearly yi/xi. The slope of the tan
gent should, therefore, be —Xi/yi. But this is exactly what the preceding
formula for the slope gives, when the equation represents a circle, i.e.
when A = B.
EXERCISES
1. Show from the result of the last article that at the points where the
curve Ax^ + By^ +(7=0 {ABC =^ 0) crosses the ?/axis its tangents are
horizontal ; and that at the points where it crosses the a;axis its tangents
are vertical.
2. Find the equation of the tangent to each of the following curves at
the point indicated. Check the result by sketching the curve carefully
and drawing the tangent from its equation,
(a) 4 a:2 4 y2 = 25 at (2, 3). (6) x^ + iy^ = S&t (2, 1).
(c) 3 x2 + 4 1/2 ::^ 16 at (2, 1).
X, § 181] IMPLICIT QUADRATIC FUNCTIONS 275
181. The Case i4 > 0, 5 < 0. We may always write the
equation (9) so that A is positive. The case where A and B
have unlike signs leads to a new type of graph.
The Graph of x^ — ]p = 9. In seeking the graph of this
equation, we observe first the following facts :
(1) The graph crosses the a;axis at the points (3, 0) and
(—3, 0), and does not cross the yaxis. Why ?
(2) The curve is symmetric with respect to both axes. For,
if the point (/i, li) is on the curve, so also is the point (A, — H).
Hence, the curve is symmetric with respect to the a>axis.
Similarly, if the point (/i, fc is on the curve, so also is the
point (— /i, Iz). Hence the curve is symmetric with respect
to the 2/axis.
(3) Solving the equation for y gives us
(16) 2/ = ± Va;2 _ 9,
This incidentally again establishes the symmetry of the curve
with respect to the xaxis. But it shows further that, if a'2<9,
y is imaginary. Hence, no part of the curve lies in the strip
of the plane between the lines x — Z and a; = — 3. In other
words all values of x between 3 and — 3 are excluded. Solv
ing the equation for x gives
a; = ± V2/' + 9.
This shows that no values of y are excluded, since 2/^ + 9 is
positive for every real value of y.
(4) The slope of the curve at the point (a^i, y^ is by § 180,
m =— •
2/1
This shows that the curve crosses the a;axis vertically, i.e. the
lines a; = 3 and aj = — 3 are tangent to the curve at (3, 0) and
(— 3, 0) respectively.
With these results in mind we now calculate the coordinates
of a few points on the curve and the slope of the curve at these
276 MATHEMATICAL ANALYSIS
points. We thus get the following table :
[X, § 181
X
3
4
5
6
y
V7
4
3V3
m
QO
fvy
f
fV3
We plot these points and those symmetrically situated with
respect to the two axes and get Fig. 150. We know from
equation (16) that y increases numerically from as a; increases
: :: :::::;: : : x:::
f 
x '
^ ~ G J2
A:^ _ ^ TLt :
"s \ r, Ay
"sZs ~ " ^ y "
_s^s_ __; ^ _ _ <L I
% \ z ^ I i :
S ^ 5 /La
\ ^ ^ ^ '
5^1, it
L S ^ Z J
\ ^ 1 ^ i
A si / t
_ j: sc dt
1 ^ V i. aL 6 'C
i zj's Jl
t ^ ^5
y /  S^ ^L
t ^ S 5
J /' S s
 ^ z : :s^^.
.^t.^ s ^^
2 2   ^s S: """
t ,Z_ \i s
z:? _ _ s s
^/   ^^
z_  . ^
i
Fig. 150
numerically from 3. We have already seen that the curve
consists of two branches. It remains only to consider what the
character of the curve is for numerically large values of x.
Equation (16) tells us that y increases numerically without
limit, as x increases indefinitely in absolute value ; i.e. the curve
recedes indefinitely from both axes. It recedes, however, in a
very definite way. For, consider the slope m of the curve at
any point (aji, 2/1). From § 180 we have, for A = l, J5 = — 1,
X, § 181] IMPLICIT QUADRATIC FUNCTIONS 277
m=^= ^
Vi ± Va^i^  9
the upper sign being used if yi is positive ; the lower, if y^ is
negative. To fix ideas, let (x^, yi) be a point in the first quad
rant and let it move out along the curve indefinitely. We de
sire to see what happens to the slope of the curve under this
condition ; i.e. when Xi becomes indefinitely large. To this end
we write m in a more convenient form, as follows :
2/ 2/] ={^^i)y
which shows that as Xi increases indefinitely, m approaches
more and more nearly the value  1. This shows that the
further the point {x^, y^) travels out along the curve in the first
quadrant, the more nearly does the direction of its motion
make an angle of 45° with the icaxis.
Consider now the equation of the tangent to the curve at
the point {xi, ?/i) :
2/2/] =
2/1
or,
^1^  2/i2/ = ^i^  Vi^
or.
This may be written
xi 9
y =— ' X
2/1 2/1
As Xi and yi become indefinitely large, the slope Xi/yi, as we
have seen, approaches the value + 1, while the term 9/2/i evi
dently approaches the value 0. Therefore, the tangent to the
278 MATHEMATICAL ANALYSIS [X, § 181
curve at the point {xi, 2/1) approaches the line
y=:x.
A line, which is the limiting position which the tangent to a
curve approaches, as the point of contact recedes indefinitely
along an infinite branch of the curve, is called an asymptote of
(lie curve.
If the point (x^, y^ recedes indefinitely along the curve in
the third quadrant {x^ < 0, 2/1 < 0), the slope is positive and the
tangent approaches the same limiting position as before,
namely, y = x. Similar considerations (or the symmetry of the
curve) show that the line
y = x]
is also an asymptote. The two asymptotes are also shown in
the figure as they are a great help in drawing the curve.
The Graph of x"^ — y"^
If, in place of the 9 in the
equation x"^ — y'^=^ just considered, we have any other positive
number, say a^, the discussion is very similar and accordingly we
can be brief. The curve of the equation x'^—y'^ = a^ crosses the pr
axis at the points (a, 0) and (—a, 0),
and does not cross the 2/axis. It
is symmetric with respect to both
axes. We have y = ± y/x^ — a^
and m = Xi/y^. We find also
1
m= ±
Fig. 151
4
from which we conclude that the curve approaches indefinitely
near the straight lines y—x and y=—x. The curve is, then, as
drawn in Fig. 151.
X, § 181] IMPLICIT QUADRATIC FUNCTIONS
279
The General Case, when C is Negative. Any equation
of the form
where A is positive and B and C are both negative, may now
be treated without much difficulty. Any such equation can be
written in the form
(17) x^  ny = a\
From this we obtain
n
a\
Fig. 152
But this shows at once, by com
parison with the last equation
considered, that the ordinates of
points on the curve x^—n^if^d?
are to the corresponding ordinates of the curve ^ — 'ip =. o? as
X/n is to 1. In Fig. 152 we have drawn both the curve
a;2 — 2/2 == d? and the curve x"^ — ^y"^ = a^, the ordinates of the
latter being just one half of the corresponding ordinates of the
former. The asymptotes of the latter are the lines y^^x and
y = ^x.
Since the asymptotes are a great help in sketching the curve,
we should have a means of obtaining their equations quickly
from the equation of the curve. From the result of § 180
(A=l, B=— n^) and considerations similar to those used in
the discussion of x'^—y^=9, we find the equations m. the
asymptotes to be
y = X and y = x,
n n
OT X— ny = and x\ny = 0. But these equations are found
by placing equal to zero each of the factors of the lefthand
member of the equation of the curve x^ — n^y^ = a^.
280
MATHEMATICAL ANALYSIS
[X, § 181
An example will show how these various results may be applied in
sketching a curve whose equation is of the form considered. To sketch
the graph of 4iX^ — 9y' = 36, we draw
first tlie asymptotes 2x—3y=0 and
2 X + 3 y = (Fig. 153). We next
place y = 0,in the given equation and
find the a;intercepts to be x = 3 and
£c = — 3. We can now sketch the
curve with considerable accuracy, since
we know what its general charac
teristics are.
""■•Jnxrr t ;lL^' "
: ■ hi: :: ± :; : _;<? :
S S T , 4^ , 
4 ^ ^s ( ^^ / 
:_::±:j:___± i^.zzt'.:.' :
:i::::±:±::i^?.±{3. .±
J /:^5_^ \
::::::±^i^::.::::::!sU:±:i::
^5"^:::::::::::^^^
/''^(^ _   ■^s*>. lX
4Ur<Tl T>T*sj I
Fig. 153
The graph of any equation of the form
^2 _ „2^2 3= ^2 {^n ^ 0, a ^ 0)
is a curve called a hyperbola. We have seen that it consists
of two branches ; it is symmetric with respect to each of
two lines, which are called the axes of the curve. One' of
these cuts the curve in two points and is called the trans
verse axis ; the other axis does not meet the curve at all. The
intersection of the axes of the curve is called the center of the
curve. The branches of the curve extend indefinitely and
approach two straight lines, the asymptotes of the curve,
which pass through the center.
We may now complete the discussion of the graph of any
equation of the form Ax^ \ By^ \ C = 0, under the hypothesis
that A is positive and B negative. We have already disposed
of the case < 0, by considering the form x^ — 7iy — a,\ The
case C > leads similarly to the form x^—n^y'^=—a^. By
interchanging x and y this reduces to the form n^x"^ — y^ = a^
which on division by n^ reduces to the case O < already con
sidered. The graph of an equation Ax"^ \ By"^ \ C=0, when
A is positive, B negative, and G positive, is therefore a hyper
bola with the center at the origin and with its transverse axis
coinciding with the 2/axis.
X, § 182] IMPLICIT QUADRATIC FUNCTIONS
281
^.= ^ =^^^
iii^?^i:S?":i
 _ l:^^Z
^^ ^^12] r
P^$
'^ T T l^h
Fig. 154
It
a
^^ ^^
^ » ^'^
^v y^ :_
_i ^.^^^
' =^'^ ±s. "
,«: _ i
Fig. 155
The following example will illustrate the method of sketching the
curve : Sketch the graph of ic^ — 4 ?/2+ 4 = 0. The asymptotes are x — 2 y =
and X + 2 1/ = (Fig. 154). Placing x = 0,
we find the ^/intercepts to be +1 and — 1.
Having marked the corresponding points and
drawn the asymptotes the graph is readily
drawn.
Finally, when 0=0, the equation
may be written in the form x^—7i'^y^=0.
This may be written
(x — ny){x + 7}y) = 0. This equation will be
satisfied by all points which satisfy either
x—ny — or icf7?y = 0, and by no others.
The locus of the equation is then two straight
lines passing through the origin. Figure 155
shows the locus of the equation 4 a;^— 9 2/^=0.
182. The Case il = or 5 = 0. If ^ = 0, 5 > 0, the equation
Ax^{By^\C=0 becomes By^ \ C — 0. If O is positive, there
is no graph. If C is negative, the graph consists of two lines
parallel to the £caxis. If C is zero, the graph is the x'axis.
When B—OjA>0, the graph of the equation consists similarly
of two straight lines parallel to the t^axis, if C is negative ; of
the ?/axis, if is zero ; and there is no graph, if G is positive.
EXERCISES
1. Sketch the graph of each of the following equations :
(a) x2  9 ?/2 = 16. (d) 9 x'^  16 ^2 _^ 16 = 0. (g) 3 x2  2 y^ = 6.
(&) x2~9i/2= 16. (e) 9x^ 16?/2 16 = 0. (/i) 3x2  12 = 0.
(c) x2  9 2/2 = 0. (/) 9 x2 'l6 1/ = 0. (0 3x2 + 1 = 0.
2. Give a detailed discussion of the graph of the equation x2— y^ =— 9
(analogous to the discussion of x2 — y2 _ 9 given in the text).
3. Give a detailed discussion of the graph of x^—n^y^=—a^. Prove, in
particular, that the asymptotes of this hyperbola are given by x^—7i'^y^=0.
4. Prove that no tangent to the curve x^  y^ = a^ has a slope that lies
between + 1 and — 1. Prove, in general, that no tangent to the curve
a;2 _ n22/2 = a2 (a =56 0) has a slope that lies between 1/n and — 1/n.
282 MATHEMATICAL ANALYSIS [X, § 183
III. THE FORM Ax^ \ By'' + Dx ^ Ey + C =
183. Recapitulation and Extension of Previous Results.
We have seen in the previous sections of this chapter that an
equation of one of the forms
By^ + Da; = 0,
or ^a;2 152/2 4 0=0
represents either
(a) a parabola, with vertex at the origin and axis coinciding
with the ojaxis or the 2/axis ; or,
(h) an ellipse, with center at the origin and axes coinciding
with the axes of coordinates ; or
(c) a hyperbola, with center at the origin and transverse axis
coinciding with the a;axis or the yaxis ; or
(d) two straight lines (which may coincide) ; or
(e) a single point (the point (0, 0)) ; or
(/) no locus.
If we replace xhj x — h and yhjy — k, in any of the above
forms, we know that the graph of the resulting equation is ob
tained from the graph of the original equation by moving the
latter so that the origin moves to the point {h, k) (the axes re
maining parallel to their original positions).
We may then conclude that an equation of any one of the
forms
(18) A{x  hy+ E{y  k)  0, B{y  ky+ D{x h) = 0,
or A(x  hy \B(yky\C =
represents either
(a) a parabola with vertex at the point {h, k) and axis coin
ciding with the line x — h = or the line y — k = 0; or
(b) an ellipse with center at the point (h, k) and axes coin
ciding with the lines x — h = and y — k = 0; or
X, § 183] IMPLICIT QUADRATIC FUNCTIONS
283
(c) a hyperbola with center at the point (/i, k) and transverse
axis coinciding with the line a;— /i=0, or the line ?/— A:=0; or
(fZ) two straight lines (which may coincide), or
(e) a single point (the point (h, k)) ; or
(/) no locus.
Now, any equation of the form
(19) Ax' \Btf + Dx + Ey\C=0
can be put in one of the forms (18) by completing the squares.
The following examples show how this may be done.
axis'" ^
'^.
Fig. 156
Example 1. Discuss and sketch the graph of y2_2yf2x + 7 = 0.
This equation may be written in the form
y2_2y=2x7,
or
l/2_2y + l=2x7 + l,
i.e.
It is accordingly a parabola with vertex at (  3, 1 ) and axis y = 1. The
graph is given in Fig. 156.
Example 2. Discuss and sketch the graph
of a;2 + y2 _ 4 a; _ 6 y + 9 = 0.
This equation may be w^ritten in the form
(x24x + 4) + (2/26?y + 9) =9 + 4 + 9,
or
(a;2)2 + (?/3)2 = 4.
Therefore the given equation represents a
circle with center at (2, 3) and radius equal
to 2. (See Fig. 157.) • Fig. 167
1, C _ 7
, i''''^:
Hi" ? ^ ■? 1 ^
284
MATHEMATICAL ANALYSIS
[X, § 183
Example 3. Discuss and sketch the graph of 9x^ + IQy'^ — ISx
+ 64 2/  8 = 0.
This equation may be written in the form
9(x2_2x+ )+16(?/2 + 4?/+ )=8,
or
9(a;22a:+l)416(?/2+4?/ + 4) =8+9 + 64=81,
i,e,
9(xl)2+16(?/ + 2)2 = 81.
 ■ ■ i, ...,.
ffffiMffl™
::: i::f :::q:: :::::
S 41  T  ( 
\ ± iL /  :
"a ±i
T "'rr^
S  It   .
3 It :
± it
Fig. 158
Hence this equation represents an ellipse whose center is at (1, — 2)
and whose axes coincide with the lines x = l,y = — 2. The remainder of
the discussion is left as an exercise. The graph is given in Fig. 158.
Example 4, Discuss and sketch the graph
of 9x236x4?/2 + 24?/ = 36.
This equation may be written in the form
9(x  2)2 _ 4(y  3)2 = 36,
which is a hyperbola whose center is at (2, 3)
(Fig. 159). It is left as an exercise to com
plete the discussion and prove that the equa
tions of the asymptotes are 3(x — 2) +
Fig. 159 2(y  3 ) = and 3(a;  2)  2{y  3) = 0.
■ ■sry' ■ ■ " ^11.
^  
5 " v" A' '
■I /ti
: ::: : s:::: ni; : :::
if / li
j/ N \
^ j^. ....
1
:::. ::. :±.: _:±:::;__:
EXERCISES
Discuss and sketch the graph of each of the following equations
1. a;2 + 42/ + 4 = 0.
2. x2 + ?/2 + 4 X  8 y + 1 = 0.
3. x2  ?/2 + 2 X = 0.
4. x2  4 X + 2/2 + 2 ?/ + 1 = 0.
5. x2 + 4 X + 2 2/2 + 4 y + 1 = a.
6. 9 x2 + 4 2/2  36 X  8 2/ + 4 = 0.
7. 9x242/2 36x + 82/ = 4.
8. 2/2 + 22/ 12x 11 =0.
9. x2 + 15 2/2 + 4 X + 60 ?/ + 15 = 0.
10. x2  3 2/2  2 X  6 2/ + 7 = 0.
X, § 184] IMPLICIT QUADRATIC FUNCTIONS 285
184. The Slope of the Curve Ax^ j^ By'^ \Dx {Ey + C = 0.
Let P(aJi, 2/i) ^^^d Q(Xi  ^x, y^ + A?/) be any two points on
the curve. Then
Ax,^ + By,^ + Dx, + ^2/i + C = 0,
^(a^i h Ax)2 + 5(2/1 + Ay)2+ i>(a;, + Aa;) + E{y, + A^/) + C = 0.
Expanding the second of these equations and subtracting the
first from it, we have
(2 Ax, + A^x + B)^x I (2 %i f SAt/ 4^) a?/ = 0.
Therefore the change ratio, or the slope, of the secant PQ, is
A^ _ _ 2 Ax^ \ A^.x + D
A.i~ 2 By, \ B^y \ e'
If we let Aa; approach zero. A?/ will approach zero also. Why ?
Therefore the slope of the curve at any point (Xi, y,) is
2Ax,\D
m = —^ — •
2By, + E
Example. Find the equations of the tangent and the normal to the
curve x^ + 4y'^ — ix\2y — S = 3i,t the point (1, 1).
Solution : The slope of the tangent at any point (xi, yi) is
8^1+2
At the point (1, 1) this slope is . Therefore the equation of the tangent
isy— l=i^(x— 1) and the equation of the normal is y — 1 = — 5(x — 1).
EXERCISES
1. Find the slope of the tangent to each of the following curves at
the point specified.
(a) x2 + 2?/ 3 = at (1, l)j
(b) x^\y^4 = at (1, V3)^
(c) a;22 2/2 + 5 = at (1, V3);
(d) 4x^\y^2xSylO = at (2, 1).
2. Find the equation of the tangent to each of the curves of Ex. 1, at
the point specified.
286
MATHEMATICAL ANALYSIS
[X, § 185
IV. THE FORM Fxy + Dx + Ey + C =
185. The Graph of xy = a. The graph of the curve
xy = a
is symmetric with respect to the origin ; for, if the coordinates
Qi, k) satisfy the equation, the coordinates {— h, —7c) also
satisfy it. Since y = a/x, it is evident that x may assume all
Fig. 160
Fig. 161
values except 0. (See § 36.) As x increases numerically
without limit, the curve approaches the line y = 0, i.e. y =0 is
an asymptote. Similarly as y increases without limit, the
curve approaches the line a? = as an asymptote. It will be
proved later that the curve is a hyperbola, provided a is not
equal to zero. If a is positive, the graph is as in Fig. 160. If
a is negative, the graph is as in Fig. 161. If a is zero, the
graph consists of the two axes x = and y = 0.
186. The Graph of Fxy \ Dx \ Ey + C = 0. If in the equa
tion xy = a we replace x hy x — h and yhjy — k, we know
that the graph of the resulting equation is obtained from the
graph of the original equation by moving the latter so that the
origin moves to the point (^, fc), the axes remaining parallel to
X, § 187] IMPLICIT QUADRATIC FUNCTIONS
287
their original positions. It follows that the equation
{x — h){y — k) = a{a^O)
represents a hyperbola whose asymptotes are x = h, y = k.
If a = 0, the equation represents the two lines x =^ h, y = k.
Example. Discuss and sketch the
graph of xy + 4x {2y = 1.
First we write
(ix±?){y±?) = l.
Then from inspection we see that the
given equation may be written in the form
(x + 2)(2/ + 4)=9.
That is, the graph is a hyperbola whose
asymptotes are x = — 2, y = — 4. (See
Fig. 162.) Fig. 1G2
187. The Slope of the Curve Fxy + Dx ^ Ey + C = 0. It
is left as an exercise to show that the slope of the curve
Fxy^Dx{Ey\C=0
at any point (x^, y^ is
Fy,AD
m = — p—  •
Fxi + E
EXERCISES
1. Discuss and draw the graph of each of the following curves :
(a) xy = l; (b) xy=l; (c) xy = 2; (d) xy = 2;
2. Discuss and draw the graph of each of the following curves.
(a) xy + 2x = S; (b) xy + 2 x \ iy = S; (c) xy  4x + Sy =2,
3. Draw the family of curves xy = a, taking several positive and
several negative values of a. How does xy = 0, compare with these ?
4. Show that any equation of the form
^ cx + d
can be reduced to the form given in § 186.
288
MATHEMATICAL ANALYSIS
[X, § 188
V. THE GENERAL FORM Ax^ + Fxy + By^ + Dx +Ey \ C =
188. The Graph. Methods of drawing the graph of an
equation in the above form will be illustrated by means of the
following examples.
Example 1. Discuss and sketch the graph of
x^ + 2xy + y'^ 2x2=0.
Solving for y, we have y = — x ±y/'2x \ 2. All values of x less than
— 1 must be excluded, for these values make 2 x + 2 negative. Similarly,
since x=—(y—l)± V — 2?/ + 3, it follows that all values of y greater than
I must be excluded ; for these values make
— 2 2/ + 3 negative. The a;intercepts are
the roots of the equation a: — 2 aj — 2 = 0,
i.e. 1 ± y/S. The ?/iritercepts are the roots
of the equation y^—2=0, i.e. ± V2. From
y —— X ± V2 X { 2 it is seen that x may
start with the value — 1 and increase
without limit. Similarly from x = — (y—1)
±V— 2y + S we see that y may start with
the value f and decrease without limit.
Using the above data and plotting the
points
Fig. 163
X
1
1
2
1±V3
y
1
±V2
1,3
2±V0
we obtain the graph in Fig. 163.
This, problem may be approached from an
entirely different standpoint. Suppose we let
y' = ± y/2 xi2 and y'f = —x. Plotting these
curves* (Fig. 164), adding the ordinates of
y' =±y/2x \2 to the ordinates ofy" = — x,
gives us the desired graph. This may be
done graphically. We have here a shear of
y' HZ t v/2 x + 2 with respect to the line y" =
Y
,CS^ it
4^
s \l
\ ^D'^^
c ^^
lS
^^ J
^ ^^
^^ Sv
^^ ^
\"^>sS
v^5^
^ ^
\ ^
3_
_r
Fig. 164
X. (See § 90.)
* Observe that the equation ?/' =± v'2 a: + 2 is equivalent to y"^ = 2{x + 1),
X, § 188] IMPLICIT QUADRATIC FUNCTIONS
289
Example 2. Discuss and sketch the
graph of
2/2  2 xy + 2 x2 — 5 a; + 4 = 0.
Solving for ?/, we have
y = x± V— x^ \ dx — 4:.
Hence, we merely have to shear the circle
1/ ±V(x l)(4x),
X + y2 _ 5 a; + 4 = 0,
with respect to the line y" = a: in order to
obtain the desired result. (See Fig. 165.)
The complete discussion is left as an exercise.
Example 3. Discuss and sketch the graph of
7 a2 + 36 xy S6y^25 =
Solving for y, we have
y = lx± iVl6a;225,
Y. ^'^N ,'
X ^'^
Z .2
y Z i
.^Z I
t7^ J^
^^v
y^ Z^^
z2^^ \
/  _ __ _
/O X
^ \ J
^^^^
Fig. 165
T7
1
I
J
/ ^
^  ^ "
U A y> .
^>  t ,^^%r^
^>s ^%^^r^'i^
^ N / X '^^^ iC
^ SZ ^ !&^
::::::_::n^^::::^^:^: :_:_::::
::^:::^:^^:SJ4^^^p
~ ) .^^"^ ^^'■' , *U.
_ 2ES'' 1 N,^.
<^ N 5^^».
yT^^i ^^l '
^^^Jl^\r •' ^ss
^^ ^  t ^5^
^ J Si
?' ' it
^z :
^2 
1
Fig. !()(>
which shows that the desired graph may be obtained by shearing
I.e.
y=±iVl6x225,
16a;2_36i/2_25 = 0,
with respect to the line y = \x (See Fig. 166.) The complete discussion
is left as an exercise.
290 MATHEMATICAL ANALYSIS [X, § 188
EXERCISES
Discuss and sketch the graph of each of the following equations :
1. Ax"^ + y"^ — i xy  X \ S z= 0. 4. iy^  ixy + x:^ = 1  x.
2. 7f2xy + 3x = 2. 5. Qy'^ 12xy \'Sx^ + ox = 6.
3. 2/28 xy +iex'^=l x2. 6. y^  6xy \ Sx^  \0x  25 = 0.
189. The Slope of the Curve Ax^ + By"^ + Fxy \ Dx + Ey
+ C = 0. It is left as an exercise to prove that the slope m at
any point {xi, y^) is
. 2By,^Fx,+E
EXERCISES
Find the equations of the tangent and the normal to each of the follow
ing curves at the points indicated.
1. 48 x2  11 a;y  17 2/2  129 a: + 24 ?/ + 81 = ; (2, 1), (3,  3).
2. x?/ 4 2 X  a;2 4 ?/2 + 6 2/ = ; (0, 0), (0,  6).
3. 81 y2 + 72 xy + 16 x2  96 x = 378 y  lU ; (3, 2).
190. A General Theorem. The results of the examples and exercises of
§ 188 suggest that the graphs of equations of the second degree involving an
xyterm are similar to the graphs of equations of the second degree in which
the xyterm is lacking. We may now prove that this is a fact. The
theorem is as follows :
Any equation of the form Ax^ + Fxy + By + Dx {■ Ey + C = repre
sents either an ellipse, or a hyperbola, or a parabola, or two straight lines
{which may coincide), or a single point, or no locus.
We shall prove this theorem by showing that if the locus of the
equation
(20) Jx2 + Fxy { By"^ + Dx + Ey \ C=0
be rotated about the origin through sl properly chosen angle d, its equation
will be of the form
(21) ^'x2 + Bhf + D'x + ^'y + C = 0.
The theorem then follows from § 183.
X, § 190] IMPLICIT QUADRATIC FUNCTIONS 291
We saw in § 137 that, if any point P(x, y) be rotated about the origin
through an angle ^ to a new position P'(a;', y'), the coordinates of P and
P' are connected by the relations :
x' =x cos d — y sin 6,
xsin d \ y cos d.
(22)
Solving these equations for x and y in terms of x', y\ we obtain
X = x' cos d \y' sin d,
^ ^ y =— x' sin 6 \ y' cos 6.
If P(x, y) satisfies equation (20), P'(x', y') will satisfy the equation ob
tained by substituting the values of x, y from (23) in equation (20) .
The result of this substitution is as follows :
A (x' cos d \ y' sin d)^+ F(x' cos^ + y'sin e){—x' sin d { y' cos 6)
+ B(— x' sin e + y' cos ey
+ D{x' cos d + y' sin d)
+ E(—x' sin e + J/' cos ^) + O = 0.
When expanded and rearranged according to the terms in x', y\ we
obtain
(24) A'x'^ + F'x'y' + B'yi'^ + D'x' + i^'?/' + C = 0,
where A' = A cos^ + P sin2 ^  P sin ^ cos 6.
F' = 2(A B) sin ^ cos ^ + P(cos2 6  sin^ ^).
B' = A sin2 ^ + P cos2 ^ + P sin 6 cos ^.
D' = Bcos e— E sin ^.
E' = D sin ^ + P cos ^.
C" = C.
Equation (24) will be of the desired form (21), if the angle 6 is so chosen
that F' = 0. Now, F may be written
(25) P' = (^P)sin2^ + Pcos2^:
F will, therefore, be equal to zero, if
tan 2 ^ ^
BA
A value of 6 satisfying the condition (26) can then always be found.*
This completes the proof of the theorem.
The following exercises will illustrate the above proof. The method
may also be used to draw the graphs of equations involving the xyterm.
* If ^ = ^, we take 26 = 90°, i.e. d = 45°.
292
MATHEMATICAL ANALYSIS
[X, § 190
EXERCISES
Determine the angle ^through which the loci of the following equations
must be rotated in order that their new equations shall contain no xyterm.
Determine the new equation and use it to draw the locus of the original.
1. Sx^ + 4xy \5y'^36 = 0.
Solution : After substituting x =x' cos \ y' sin ^, y z=— x' sin d
4 y' cos 6, the equation becomes
(1) (8 cos2 ^45 sin2 6 4: sin 6 cos e)x'^
+ [6 sin d cos ei 4(cos2 d — sin^ e)'\x'y'
+ (8 sin2 6+ b cos^ ^ + 4 sin ^ cos 0)i/'2 _ 35 _ q.
2 tan e
Therefore,
tan 2 ^ = —
3 1  tan2 d
Solving this equation for tan 6, we have
4 tan2 ^ _ 6 tan — 4 = 0,
or tan ^ = 2 or — ^.
We choose tan 6
rant) ; therefore
2
V5'
2 (^ in first quad
.
5 '
. "T^ ^
._ ^ 
^s ^
/ \"
S ^"^
t ^'
S.^J'
jj ^
^^
ji:_.i
\^
\ 1 /
^'\
\ J^
^^ A
^ mrcltan 2
"■ \ y'
\
.
Sin
cos d .
1
Substitutin,
obtain
2. x2  ?/2 __ 2 xy  12 = 0.
5 these values in (1) we
4 x'2 + 9 ?/'2 = 36.
The desired graph is obtained from the
grajjh of this equation by rotating it
through the angle — Q about the origin.
The construction of the adjacent figure
explains itself.
5. 3 x2  2 a;?/ + 1/2 ^ 6 = 0.
12 = 0.
6. 8x2
12 xy + 3 y2 _ 36 = 0.
3 ?/2 + 42 = 0.
3. x2y2_.2x?/ + 2x
4. xy = 4. 7. 2 x2  12 xy
8. 6 x2 + 4 xy  ?/2 + 48 X  12 y  10 = 0.
9. 9 2/2 + a;2 + 2 xy = 0.
10. Prove that the locus ot xy = c may be rotated about the origin so
as to coincide with the locus of x^—y'^ — a^, provided a^ =±2c.
11. With the notation of § 190, prove that A' \ B' = A \ B and that
{A'  B')^ + F'^ =(A BY + F\
PART III. APPLICATIONS TO GEOMETRY
CHAPTER XI
THE STRAIGHT LINE
191. Introduction. We have hitherto used coordinates pri
marily for the purpose of representing functions graphically
and investigating the properties of those functions. We have
seen that every continuous function defines a curve or a
straight line, the graph of the function. Thus far, we have
laid emphasis only on the discovery of the characteristics of the
functions from the known properties of the curves that repre
sent them.
Conversely, we have seen that every curve or straight line,
in the plane of a system of rectangular coordinates, defines a
function ; i.e. the points of any such curve associate with every
value of X one or more values of y. If this function can be
determined when the curve is given, the properties of the
curve may be studied from the properties of the function.
This function is usually expressed by means of an equation in
X and y, called the equation of the curve. We propose now to
study the properties of various curves by means of their equa
tions. (See § 62.)
Up to this time, we have used different scales on the two
axes whenever it was convenient to do so. Throughout this and
the next four chapters we shall assume, unless the contrary is
specifically stated^ that the units on the x and yaxes are equal.
294
MATHEMATICAL ANALYSIS [XI, § 192
192. The Distance between two Points. Given the two
points Pi {xi, yi) and P2 {x2, y^)^ let us find the length of the
segment PiP^ If a line be drawn through Pj parallel to the
icaxis and another through P2 parallel to the 2/axis to form
the right triangle P1QP2 (Fig. 167), we have at once
(1)
P,P,=^J\Q' + QPl
T
P.
^
N,
Pr^
Q
Mr
M,'X
Fig.
Pi
M,
N,
JU
The segment PiQ is equal to the projection M^M^ of PxPi on
the a^axis and (^P^ is equal to the projection NiN^ of P1P2 on
the 2/axis. By the result of § 37, we have
PiQ= M^M^ = x.^— ccj,
QA=i^==2/22/i.
Substituting these values in (1), we have the desired formula :
(2)
PxP,=V(x2Xiyj{y,yi)\
193. The Simple Ratio. Given two distinct points Pj, P2
and any point P (distinct from P2) on the line P1P2, the ratio
P1P/PP2 is called the simple ratio of P with respect to Pi, P2.
The linesegments in this definition are directed segments.
Accordingly the simple ratio of P with respect to Pi, P2 is
positive if P is between Pi and P2, and negative if P is on
either prolongation of the segment P1P2.
XI, § 194]
THE STRAIGHT LINE
295
194. Point of Division. The coordinates (x, y) of the point
P on the line joining Pi (x^, y^ to P2{x2y 2/2) such that the
simple ratio
are given by the formulas
(3)
;C=£l±J^, ^^J/l + Xyg.
1 + X
l + X
Proof. Draw lines through Pi, Pg,
P parallel to the axes, meeting the
jcaxis in Mi, M^, M, and the yaxis
in Ni, N2, N, respectively (Fig. 168).
Then, since P1P/PP2 = X, we have
MiM ^ ^^=X
MM^ ' NN,
The first of these relations gives (by § 37)
X2 — X
Solving this equation for x gives
Xi + \X2
r
N
Ml /
P
M
■^2
/
X
Pi N,
Q
Fig. 168
x =
1+A
Similarly from the second relation above we obtain
y =
_ 2/1 + ^2/2
1+X
The midpoint of P1P2 is obtained from the value A = 1. Why ?
Accordingly the coordinates of the midpoint of P1P2 are
f xi + X2 Vi + y2 \
l~2' 2 )
296 MATHEMATICAL ANALYSIS [XI, § 194
EXERCISES
1. Find the distance between the following pairs of points : (1, 2)
and (5,3); ( 1, 6) and (2, 3); (2, 1) and (1,4); (3,4).
and (1,4).
2. Find the lengths of the sides of the triangle whose vertices are
(—1, 1), (4, — 4), and (1, 3). Prove that it is a right triangle.
[Hint : A right triangle is the only kind of triangle in which the square
of one side is equal to the sum of the squares of the other two sides.]
3. An isosceles triangle has its vertex at (4, 4) and the vertex of one
of its base angles at (0, — 1). The vertex of the other base angle is on
the a;axis. Find the coordinates of the latter vertex.
[Hint : Let the unknown point be (ic, 0) and equate the equal sides.
How many solutions are there ?]
4. Find the coordinates of the point whose simple ratio with respect
to (2, 1) and (—4, 7) is 2. Find the coordinates of another point
whose simple ratio with respect to the same two given points is — 2.
Draw a figure illustrating this problem.
5. Check the result of Ex. 4 by calculating the lengths of the seg
ments involved.
6. Find the coordinates of the point which divides the segment from
(2, — 1) to (— 4, 3) internally in the ratio 1 : 4.
7. Find the coordinates of the midpoints of the sides of the triangle
in Ex. 2.
8. A quadrilateral has its vertices at the points (—2, 1), (3, 1),
(5, 3), and (0, 3). Show that its diagonals bisect each other. What
kind of a quadrilateral is it ?
9. Find the coordinates of the points of trisection of the segment
from (3, 6) to (0, 3).
10. A triangle has its vertices at the the points (0, 4), (2, — 6), (  2,
— 2) . Find the coordinates of the points two thirds of the way from
each vertex to the middle point of the opposite side, and thus show that
the three medians of the triangle all pass through the same point.
11. The vertices of a triangle are (xi, yi), {xi,y2), (scs, yi). Find
the coordinates of the point of intersection of the medians.
12. Show that the triangle ^1(4, 1), i?(l, 4), C(5, 6) is isosceles.
XI, § 194] THE STRAIGHT LINE 297
13. One end of a line whose length is 13 units is at the point (3, 8).
The ordinate of the other end is 8. What is its abcsissa ?
14. The middle point of a line is (2, 3) and one end of the line is at
the point (4, 7). What are the coordinates of the other end ?
15. The points (2, 1), (3, 4), (— 1, 7) are the midpoints of the sides
of a triangle. Find the coordinates of the vertices.
16. Find the area of the isosceles triangle whose vertices are (4, 1),
(1, 4), (5, 5) by finding the length of the base and the altitude.
17. What equation must be satisfied if the points (a:, y), (2, 1), (1, 4)
form an isosceles triangle the equal sides of which meet in (ac, y)?
18. Prove that the points ( 2,  1), (1, 0), (4, 3) and (1, 2) are the
vertices of a parallelogram.
19. The line from (xi, yi) to (^2, ?/2) is divided into 5 equal parts.
Find the coordinates of the points of division.
20. A point is equidistant from the points (2, 1) and (—2, 1) and 7
units distant from the origin. Find its coordinates.
QUESTIONS FOR DISCUSSION
1. Does the distance between two points depend on the order in
which the points are taken ? Does the formula for the distance give the
same result no matter in which order the points are taken ? Why ?
2. Does the simple ratio of a point with respect to Pi, P2 depend on
the order in which the points Pi, P2 are taken ? What is the relation
between the simple ratio of P with respect to Pi, P2 and the simple ratio
of P with respect to P2, Pi ?
[ Hint. The answer to this question follows most easily from the defini
tion of simple ratio. Prove the relation in question by means of the
formulas in § 194. ]
3. Can the simple ratio of a point P with respect to Pi, P2 be — 1 ?
Why ? As the simple ratio approaches —1 what is the motion of P?
4. What can be said of the position of the point P, if its simple ratio
with respect to Pi, P2 is positive ? if its simple ratio lies between and
— 1 ? if its simple ratio is less than — 1 ?
6. If the simple ratio of P with respect to Pi, P2 is X, what is the
simple ratio of Pi with respect to P and P2 ? of P2 with respect to Pi
and P ?
298
MATHEMATICAL ANALYSIS
[XI, § 195
195. The Area of a Triangle. One Vertex at the Origin.
Let us try to find the area of a triangle whose vertices are
0(0, 0), Pi(xi, yi), and P2(^'2j 2/2) I^^t the angles XOPi and
XOP2 be denoted by Oi and O2, respectively, and let the angle
F1OP2 of the triangle have the absolute measure 6 (Fig. 169).
Fig. 169
The area of the triangle is then equal to ^ OPi • OP2 sin 0.
Now, the directed angle P1OP2 differs from 62 — ^1, if at all,
only by multiples of 360° (§ 101). Therefore
sin ^ = ± sin (PiOP2)= ± sin {$2  ^1).
The area of the triangle OP1P2 is, then,
A=:±^OPi' OP2 sin(^2  ^1)
= ± I OPi ' OPsCsin $2 cos $1 — cos $2 sin ^1)
(§138)
= ±iOP,OP,
2/2 _^i ^ Jh\
OP2 OPi OP2 OPj
The area of the triangle OP1P2, in the ordinary sense of the
term, is therefore equal to the absolute value of the expression
l(xiy2  XiVi).
For some purposes it is convenient, however, to regard the
area enclosed by a curve as a signed quantity, just as we have
XI, § 195] THE STRAIGHT LINE 299
found it convenient to regard linesegments and angles as
signed quantities.
To this end, we observe that a point moving on the boundary
of an area may make the circuit in either of two opposite
directions (Fig. 170). With each of these directions is asso
ciated a definite rotation about a ^...,,^^ ^^^
point within the area. If the bound ( ^^X ^ J^^
ary is traversed in a direction which \S^ X^x"^
produces a positive rotation about a positive arcuU Negative circuit
point within the area, the circuit and ^^^ ^"^^
the area are regarded as positive ; if the boundary is traversed
in the opposite direction, the circuit and the area are regarded
as negative. Hence if an area is represented by a signed
number, the sign of this number tells us the direction in which
the boundary is traversed.
In case of a triangle OP1P2 (Fig. 169) the order in which
the vertices are written determines a direction of traversing
the boundary. If OP1P2 is positive, OP2P1 is negative, and
vice versa. Now in going around the triangle in the direction
OP1P2, a segment OP joining to a point P moving on P1P2
generates a directed angle PiOP^ This angle is positive or
negative according as the circuit OP1P2 is positive or negative.
Moreover the measure of the angle PiOPo differs from 62 — ^1,
if at all, only by multiples of 360°. The expression
i OP, ' OP; sin((92  ^1)
is, therefore, positive or negative according as the circuit OP1P2
is positive or negative. We have then finally :
The area of a triangle OP1P2 is given in magnitude and in
sign by the formula
(4) Area OP^P. = lixiUo  ^^2^1)
300
MATHEMATICAL ANALYSIS
[XI, § 196
196. The Area of Any Triangle. The convention as to the
sign of an area is serviceable in deriving a formula for the area
of any triangle in terms of the coordinates of its vertices.
Let the vertices be Pi(xi, yi), P2(^) 2/2)) ^3(^*3? Ik) Join these
vertices to the origin by lines OPi, OP2, OPs. We novr con
sider the three possible cases, according as the origin is inside
><a!
^^
Fig. 171
'1^
Y
X
\y
X
k
Fig. 172
Fig. 173
(Fig. 171), outside (Fig. 172), or on, a side (Fig. 173) of the
triangle PJ^JP^. Then in all cases, we have
A P,P,P, = AOP.P, + AOP2P3 + AOP.Pi,
if due regard is taken of the signs of the areas. Hence
(5) Area of A P.P^P^^ \{if',x,  x^y, + y,x^  x^y^ + y^x^  x^y,).
It might appear that this formula is difficult to apply. The following
method makes it very simple. Write the coordinates of the
vertices in two vertical columns as indicated, repeating the
coordinates of the first vertex. Multiply each x by the y in
the next row and add the products. This gives iCi?/2+X2y3+X3i/i.
Then multiply each y by the x in the next row and add the
products. This gives yiX2 + y^Xs + ysXi. Subtract the second sum from
the first and divide the result by 2. The final result will be the area
sought, with its proper sign* A similar method may be used for finding
the area of any convex polygon whose vertices are given. See Exs.
6, 7, 8, pp. 301, 302.
* The student familiar with the elements of the theory of determinants
will .observe that the area can be expressed as
A=i
Xi
yi
X2
Vi
X3
ya
Xi
y\
Xl
V\
1
Z2
2/2
1
Xg
Vs
1
XI, § 197] THE STRAIGHT LINE 301
Example. Find the area of the triangle whose vertices are Pi(— 4, 3),
P2(— 1,  2), PaC— 3,  1). We write the coordinates of the
vertices in two columns, repeating those of the first vertex.
Performing the first step described in the previous paragraph ~ q ~ i
we obtain 8 + 1 — 9 = 0; the second step gives — 8 + 6 + 4
= 7 ; the third step gives 0— 7 =— 7 ; dividing this by 2,
we obtain — 3^ as the area of triangle P1P2P3. The magnitude of the
area is 3^ square units, and the direction Pi to P2 to P3 is negative.
Draw the figure and verify the latter statement.
197. Condition for CoUinearity of three Points. If three
points Pi, P2, P3 are collinear, the area of the triangle formed
by them is zero ; conversely, if the area of a triangle is zero,
the three vertices are collinear. Therefore, a necessary and
sufficient condition that three points be collinear, is that the right
hand member of (5), p. 300 be equal to zero.
EXERCISES
1. Find the areas of the following triangles and interpret the sign of
the result in each case. Illustrate by appropriate figures.
(a) (1, 3), (4, 2), (2, 5). (c) ( 5, 2), ( 4,  3), (1, 1).
(6) (2, 4), ( 3, 1), (1,  7). (d) (a, a), ( 6,  6,) (c, d).
2. Show that the following sets of three points are collinear :
(a) (0,1), (2,5), (1, 1). (c) (1,2), (6, 1), (4, 5).
(&) (2, 1), ( 4, 4), (4, 0). (d) (0, 6), (1, a6), (a, a^b).
3. The point (h, h) is collinear with (2, 5) and (5,  3). Find its
coordinates.
4. Find the point on the yaxis collinear with (2, 5) and (5, — 3).
6. Under what conditions on a, b, c, and d are the points in Ex. 1 (d)
collinear ? Interpret each of the conditions geometrically.
6. Area of any polygon. Show that the method of § 196 may be ex
tended to derive a formula for the area of any polygon in which two
sides do not cross each other, and that if P1P2P3 " Pn are the vertices of
the polygon taken in order around the polygon, we have
Area of polygon = A OP1P2 + A OP2P3 + A OP3P4 + ... + A OP„Pi,
if due regard is paid to signs.
302 MATHEMATICAL ANALYSIS [XI, § 197
7. Find the area of the quadrilateral whose vertices are (1, 2),
(2,3), (3, 4), and (4, 5).
8. Find the area of the polygon whose vertices are (4, 1), (2, 3),
(0,4), (2,3), (4,1).
9. Prove that the points (1, 2), (3, 6), (—1, — 2) are collinear.
10. Show that the area of the triangle whose vertices are (2, 6),
(— 4, 3), (—2, 7) is four times the area of the triangle formed by join
ing the middle points of the sides.
198. Applications to the Proof of Geometric Theorems.
We shall now give a few elementary examples to show how
the methods hitherto developed may be used in the proof of
geometric theorems.
Example 1. Prove that the line joining the vertex of any right tri
angle to the midpoint of the hypotenuse is equal to half the hypotenuse.
Let ABC be any right triangle. In order to apply the methods of
coordinates we must first locate a pair of coordinate axes. Any two
perpendicular lines will serve the purpose, but the
work incident to the solution of many problems
may usually be greatly simplified if we choose
the axes judiciously. In this case it is convenient
(a,o')~x ^° choose the legs of the triangle as axes. The
F 174. coordinates of the vertices are then (Fig. 174)
(0, 0), (a, 0), and (0, &). The midpoint of the
hypotenuse is (§ 194) (a/2, b/2). The length of the line joining this
point to (0, 0) is V(a72JHjbj2y = ^VoM^ But the length of the
hypotenuse is Va'^ + b'^. This proves the theorem.
Example 2. Prove that the diagonals of a parallelogram bisect each
other.
Let ABCD be any parallelogram. Let a side of the parallelogram lie
on the Xaxis a vertex being at the origin.
(See Fig. 176.) We may assign the coordinates
{a, 0) to the vertex B, and (6, c) to the vertex
D. The coordinates of C will then be (af&, c). —q
Why?
We now calculate the coordinates of the
•PCb.c) C(a+6.o)
^
£{0,0) X
Fig. 175
midpoint of ^C and also of the midpoint of BD, by the formula of
§ 194. It will then be seen that the midpoints coincide.
XI, § 198] THE STRAIGHT LINE 303
Example 3. Prove that if the lines joining two of the vertices of
a triangle to the midpoints of the opposite sides are equal, the triangle
is isosceles.
Let ABC be the triangle, M^ N the midpoints of
the sides AC^ BC, respectively, with AN'= BM. Let
the a;axis lie along the side AB and let the yaxis
pass through the vertex C (Fig. 176). Let the coordi
nates of A, B, C he (a, 0), (&, 0), (0, c) respectively.*
We must first state the hypothesis of the theorem
analytically, i.e. in terms of the coordinates. To this ^^'
end we note that the midpoint of AC is M = {a/2, c/2), and that
BM^'
Y
/
CCo.c)
J
>
P
^ X
A(a,o)
. B(b,o)
Similarly, we have AN^ = (a^Yh
By hypothesis, AN = BM. Hence we have
■ This condition gives
(f)=.(«>
which, when simplified, gives either a = b or a=— b. The first result
would imply that the points A and B coincide, which is contrary to the
hypothesis, and is therefore rejected. The second result yields readily
that AC= BC, which was to be proved.
EXERCISES
1. Prove analytically that the diagonals of a rectangle are equal.
2. Prove analytically that the line joining the midpoints of two sides
of a triangle is half the third side.
* In the figure a is a negative number. However, the discussion that follows
applies at the outset to any numbers, a, b, c. It will appear later in the dis
cussion that, under the hypothesis of the theorem, a and b must have opposite
signs. One of the advantages of the analytic method is the fact that it is
general, and that ordinarily special cases do not have to be considered
separately.
304 MATHEMATICAL ANALYSIS [XI, § 198
3. Prove analytically that two triangles with the same base and
equal altitudes have the same area.
4. ABCD is a parallelogram, with A^ C as opposite vertices. JIf and
iVare the midpoints of the sides AB and CD respectively. Prove ana
lytically that the lines AN and CM trisect the diagonal BD.
5. If P is any point in the plane of a rectangle, prove analytically
that the sum of the squares of the distances from P to two opposite
vertices of the rectangle is equal to the sum of the squares of the dis
tances from P to the other two vertices.
6. Prove analytically that, if the diagonals of a parallelogram are
equal, the figure is a rectangle.
7. Prove analytically that the two straight lines which join the
midpoints of the opposite sides of a quadrilateral bisect each other.
8. Show analytically that the figure formed by joining the middle
points of the sides of any quadrilateral is a parallelogram.
9. If ilf is the midpoint of the side BC of any triangle ABC, prove
that AB2+ AC^ = 2(AM^ + MC'^).
10. Prove analytically that the distance between the middle points of
the nonparallel sides of a trapezoid is equal to half the sum of the
parallel sides,
11. The difference of the squares of any two sides of a triangle is equal
to the difference of the squares of their projections on the third side.
12. Prove that the sum of the squares of the sides of any quadrilateral
is equal to the sum of the squares of the diagonals plus four times the
square of the distance between the middle points of the diagonals.
13. If A^ jB, O, Z) are four points of a line prove the relation (due to
Euler) : AB ■ CD\AC ■ DB\AD • BC=0. (The segments are directed.)
14. If M and iV, respectively, are the midpoints of two segments J.B and
CD on the same line, show that 2 il/iV= AC + BD= AD + BC.
15. If 31 is the midpoint of AB and P any other point of the line AB,
show that PAPB= PM^  MAK
16. Two sources of light of intensity a and /3 are situated at the points
A and B respectively of a line. Find the position of a point on the line
which is lighted with the same intensity by the two points. How many
points satisfy the relation ?
[Hint : The intensity of light at a point varies inversely as the square
of the distance of the point from the source of light and directly as the
intensity of the source.]
XI, § 198] THE STRAIGHT LINE 305
17. Two objects of weights Wi, w^ are situated at the points A^ Ao.
The center of gravity of the two objects is defined to be the point of the
line A1A2, whose simple ratio with respect to Ai, A2 is W2/W1. If Ai, A2
are on the xaxis, and their coordinates are Xi, X2^ find* the coordinate of
the center of gravity. Show that the center of gravity does not exist, if
Wi = — W2. Give an interpretation to a negative w.
18. Given n weights lOi, wo, ■•■■,Wn situated at the points Ai, A2, ••, A^
on a line. Find the center of gravity of ^1, A2 with weights Wi, W2 ;
then the center of gravity of the point found taken with th^ weight
Wi 4 W2 and A3 with the weight W3 ; then the center of gravity of this
new point taken with "the weight Wi + W2 + 103 and A^ with the weight 104 ;
and so on. Show that when all the ?i points have been used, there is
obtained a point which is independent of the order in which the points
were taken. The point thus determined is called the center of gravity of
the n points. When does no center of gravity exist ? Under what coy
ditions is it indeterminate ? Show that if the latter conditions hold, each
of the given points is the center of gravity of the remaining ones each
taken with the weight assigned to it.
19. The first (or static) moment of a point P of weight w about a line
I is defined to be the product of lo by the distance of P from I. Given n
points Pi =CXi, yi)(i= 1, 2, .., n) in a plane with weights lOc, respec
tively, determine the coordinates of a point P of weight lOi + «?2+ •• 4 Wn
such that its moment about the xaxis shall be equal to the sum of the
moments about the xaxis of the points P, and such that its moment
about the yaxis shall be the sum of the moments about the yaxis of the
points Pi. The point P is the center of gravity of the set of points. Com
pare with the result of Ex. 18.
20. The second moment or the moment of inertia of *a point P with
respect to a line I is defined to be the product of the weight 10 of P by the
square of its distance from the line. Given n points Pi in a plane whose
distances from a fixed line I are Xi, and whose weights are Wi respectively.
Let Ml be the sum of the first moments, M2 the sum of the second
moments of these points about the line I. Let V be a second line, paral
lel to the first and h units from it (to the right or left according as h is
positive or negative), and let Mi and iHf ^ be the sum of the first and sec
ond moments of the given points about I'. Let W be the sum of
the weights Wi { 102 + ••• + Wn. Show that
M'i = MihW and M'2 = Mif2hMi + h^W.
306 MATHEMATICAL ANALYSIS PCI, § 199
199. Directed Lines and Angles. An angle from a directed
line li to a directed line I2 is an angle through which li must
be rotated to . make its direction coincide with that of Zg.
Any such angle we denote by {li y. Clearly
if li and Zg intersect in a point M (Fig. 177),
(Zi Z2) is the directed angle from Zi to l^ as
^^' ' defined in § 98 since the directions of Zj and Z2
define uniquely the halflines issuing from M. As we observed
in § 101, an angle (Zi I2) may have various determinations
diifering from each other by multiples of 360°.
The angle from the ajaxis to a directed line Z is called the
inclination of Z (Fig. 178). If the inclination of a directed line
Zi'is 61 and the inclination of a directed line l^
is $2, the angle from Zi to Z2 is given (§ 101) by
the equation
(6) (/i/2)=e2ei,
where the equality sign means equal except
possibly for multiples of 360°. Fig. 178
200. Undirected Lines and Angles. If two lines Zi and Z2
are not directed, an angle from Zi to Z2, defined as an angle
through which l^ must be rotated to make it parallel to Zg, will
have various determinations which differ by
multiples of 180° (Fig. 179). The smallest
positive (or zero) angle from Zi to Z2 is then
Fig. 179 unique and less than 180°. The inclination of
an undirected line is defined as the smallest positive (or zero)
angle through which it is necessary to rotate the a;axis in order
to make it parallel to the line. In Chapter III we used the
slope m of a line to measure its inclination. It follows almost
immediately from the definition of slope m and inclination 6
that we have m = tan 0.
XI, § 201] THE STRAIGHT LINE 307
To calculate the angle from a line l^ to a line l^ we make use
of (6), § 199, if the inclinations 0^, 6^ of l^, I2 are known. If
the slopes mj, m^ of Zj, I2 are given, we lind from (6), § 138
tan (Zi /a) = tan (0.  O,) = tan 62  tan 0^
^ ' '^ ^  /^ 1 + tan O2 tan ^1
But tan ^1 = wii and tan 0^ — ^2 Hence we have
As special cases of this relation we obtain the familiar condi
tion for parallelism and perpendicularity (§§ 64, ^^^. For, if
the lines are parallel, (Zi l^ — 0° or 180° ; hence m^ — m^.
If the lines are perpendicular, (Zi, Zo) = 90° or 270° ; hence
lfmim2 = 0, or mi= •
mo
201. Standard Forms of the Equation of a Straight Line.
We recall here for reference the standard forms of the equation
of a straight line derived in Chapter III :
The general equation : Ax \ By ^ C = 0.
The slope form : y = mx + b.
The pointslope form : y — Ui = ^{x — Xi).
The last two forms are not general, since they will not serve to
represent lines parallel to the 2/axis. The first is general. If
the first represents a line not parallel to the ?/axis (B =^ 0), it is
readily reduced to the slope form, by solving the equation for y :
A C
y = X — — •
^ B B
This yields, as was shown in § 63,
A
308 MATHEMATICAL ANALYSIS [XI, § 201
EXERCISES
1. Construct a line through the point (— 2, 3) having an inclination
of 00°. What is the slope ? Write the equation of the line. Find the
points at which the line crosses the a:axis and the yaxis.
2. Proceed as in Ex. 1 for a line passing through the point (2, — 3)
with an inclination of 135°.
3. Find, to the nearest minute, the inclination of each of the follow
ing lines. Use a table of natural functions.
(a) 2xSy = 0. (c) x = 2.1y + 3.5. (e) x  ?/ + 249 = 0.
{b) y = OAx{l.'I. (d) 7x\ Sy8 = 0. (f)x + 2y+6 = 0.
4. Fhid the tangent of the angle from the first line to the second line
of each of the following pairs. Then find the angle.
{a)2xSy = 0, (c)x + 3y3=0,
ic + 2y+7=0. 3x — ?/ + 6=0.
(b) 6x + 2y 10 = 0, (d) y = 2x + 3,
2x32/ + 6 = 0. Sx + y6 = 0.
5. Find the equation of the line through (4, 5) and parallel to the
line joining (—1,2) and (2, — 3).
6. Find the equation of a line through the intersection of2x + y — 5=0
and x — 3y+5 = 0, and perpendicular to the line 2x — 3?/fG = 0.
7. An isosceles triangle has for its base the line x—2 y+2=0 and for
its vertex the point ( — 3, 5). The base angles are 45°. Find the equations
of the other two sides and the coordinates of the other two vertices.
8. Given the lines aiX + biy + ci = and a^x + 62?/ { C2 = 0. Show
that they are parallel, if and only if 05162 — «26i = ; and that they are
perpendicular, if and only if aia2 + &162 = 0.
9. The sides of a triangle have slopes equal to ^, 1, and 2. Show
that the triangle is isosceles.
10. Find the angles of the triangle whose vertices are (3, 4), (— 3, 6),
and (2,  1).
11. Find the slope of the bisector of the angle which a line of slope — 2
makes with a line of slope 3.
12. The slope of a line AB is 2. Find the equation of a line through
the origin which makes with AB an angle whose tangent is — 1.
13. P is any point on the curve whose equation is 2/2 = 4 x. Show that
the tangent to the curve at P bisects the angle which the line joining P to
the point (1, 0) makes with the line through P and parallel to the xaxis.
XI, § 202] THE STRAIGHT LINE 309
202. The Expression Axi{Byi\ C. The expression x — 2y
f 3 has the value + 2 when x = l and y = 1', the value when
x = —l and 2/ = 1 ; the value — 2 when a; = — 3 and y=l.
The only interpretation we are able thus far to give to these
facts is that the second set of values for x and y are the coordi
nates of a point (—1, 1) which is on the line whose equation is
x{2y { 3 =0, while the other sets of values are the coordi
nates of points not on this line.
It seems reasonable to expect, however, that the value of
the expression xi—2yi\3, where (x^, y^) is any point in the
plane, must have some relation to the line whose equation is
x — 2y\3 =0. This relation is indeed very simple. * The
reader should have no difficulty in proving that the value f 2
obtained ^bove from the point (1, 1) represents in sign and
magnitude the directed segment drawn parallel to the a;axis
from the line to the point (1, 1). Similarly, the value — 2
represents the segment drawn parallel to the a>axis from the
line to the point (—3, 1).
We proceed to show that a similar result applies to the
values of the lefthand member of the equation of any line in
the form Ax^By\C=0.
Let the line I (Fig. 180) be the line whose equation is
Ax\By{ (7=0, where we assume A=^0,
and suppose the equation has been
written so that A is positive. Why is
this last always possible? The line is
then not parallel to the ajaxis. Why ?
Let Pi{xi, yi) be any point in the plane
and let Q(/i, y^ be the point in which
the line through P parallel to the a;axis meets I. Since Q is
on ?, we have ^i , td , n a
or By^\C =  Ah.
e;^!^ Pi (x,.v,)
Fig. 180
310 MATHEMATICAL ANALYSIS [XI, § 202
The value of Axi { Byi + C, wliicli we are seeking, is therefore
equal to Axi — Ah, or A{xi — h). But Xi — h represents, in sign
and magnitude, the segment QPi We have then,
Ax^ + By,+ C = A'QP;.
We conclude that, HA is positive, the number Axi f Byi + C is
positive if {xi, 2/1) is to the right of the line Ax\By{C=0, and
negative if (xi, 2/1) is to the left of this line. Moreover, Axi + Byi + C
is proportional to the horizontal distance from the line to the point
(a^i, 2/i)
Finally, if ^ = and B =^ 0, we may suppose the equa
tion By{C=0 so written that B is positive. The line I is
then parallel to the a;axis. Writing its equation in the form
y =— C/B, it is readily seen that the expression
m
2/. (t5^ 1 = 2/1+ J
represents the directed segment drawn parallel to the yasia
from the line to the point Pi (Fig. 181). We may then con
clude that, B being positive, the number Byi^O
^li^vVi) is positive if the point (x^, y^) is above the line
1 By + C = 0, and negative if the point {x^, y^
is below this line. Moreover, Byi + C is jyro
portional to the distance of the point from the
line.
Fig. 181 ^ , ,. , ,. . . ,
By the preceding results, we may distinguish
between the positive and negative sides of a line. If the equa
tion of a line is written in the form Ax \ By { C = and so
that its first term is positive, the righthand side of the line is
positive and the lefthand side is negative, unless the line is
parallel to the ajaxis. In the latter case the upper side is
positive and the lower side is negative.
XI, § 203]
THE STRAIGHT LINE
311
T
/;
qA^t,,.,
/
/
/^
X
Fig. 182
203. The Distance of a Point from a Line. The results of
tlie last article enable us to find the perpendicular distance
of a point Pi(aJi, y^ from the line whose
equation is Ax + By f (7 = 0. If
A 4=^, the required distance d = MPi
(Eig. 182) is evidently equal to QPi sin^,
where 6 is the inclination of the line.
This is true whether the inclination is
acute or obtuse, and whether P^ is on
the positive or negative side of the
given line. Since 0° ^ ^ < 180°, sin 6 is necessarily positive, and
d = QP^ sin 6 will have the same sign as QPi ; i.e. it will be
positive when Pi is on the positive side of the line, and nega
tive when Pi is on the negative side.
We have, from the preceding article,
QP _ Axi + By, + C
and, since tan B = — A/B, we have
sin^ = ^
VA^{&
Hence, the required distance is
(8) MPi=d = '^^'i±^Mi±£,
If .d = 0, the required distance, by § 202, is simply
, , C Byi + C
^^ B B
But this is precisely what (8) becomes for ^ = 0. Hence (8) is
true in every case.
The distance d is positive if (xi, y^ is on the positive side of
the line, and negative if (a^i, 2/1) is on the negative side, provided
312
MATHEMATICAL ANALYSIS [XI, § 203
the equation is written in the standard form with the first term
positive.
Example 1. To find the distance from the line 2a;— S?/— 10 =0 to
the point (— 3, 1). Since the equation is in standard form the desired
result is obtained by substituting the coordinates of the given point in the
lefthand member of the equation and dividing by the square root of the
sum of the squares of the coefficients of x and y. Hence the distance d
^® ^^ 2(3)5. 110 ^21
V2'^(5)2 V29
The negative sign indicates that the point is at the left of the line.
Example 2. Find the equation of the bisector of the acute
+ 12 = and 4 ic  3 ?/ + 6 = 0.
First draw the lines (Fig. 183).
know from geometry that the bisector of
an angle is the locus of the points equidis
tant from the sides of the angle. Let (x, y)
be any point on the desired bisector. In
spection of the figure shows that (aj, y) is
on the positive side of one of the lines and
on the negative side of the other. Hence,
any point on the desired bisector must
satisfy the condition that its distance from
one of the lines is equal to minus its dis
tance from the other. This condition is
expressed by the equation :
between the lines 3 x
4y
p
T7
A
/.
/
/y
y
p'
J
^^
/
^^'/
<^J
7
 ,'
X
^i
f
 i^i
7
■■
angle
We
Fio. 183
(9)
or
(10)
3 X — 4 j/ f 12
5
4 a  3 y h 6
5
18=0.
7x7y
Moreover, any point which satisfies relation (9) is a point of the bisector.
Hence, we conclude that the equation 7x — 72/ + 18 = Ois the required
equation.
NoTK. Had the equation of the bisector of the obtuse angle been
desired the figure shows that in this case a point on the bisector is either
on the positive side of both lines or on the negative side of both lines.
Hence, any such point must satisfy the relation obtained by placing its
distance from one line equal to its distance from the other line. The
equation of this bisector is x f y 4 6 = 0.
XI, § 203] THE STRAIGHT LINE 313
Example 3. Prove that the locus of a point which moves so that the
algebraic sum of its distances from any number of fixed lines is constant,
is a straight line.
Each of the given straight lines has an equation of the form
ax+by \ c = 0. The distance of any point (x, y) from such a line is
ax \ by ■^ c ^
Va^ + &2
The equation of the required locus is, therefore, of the form
aiX hbiy \ ci ^ ^ a nX + M + Cn _ q^
Since this is an equation of the first degree, the locus is a straight line.
EXERCISES
1. Without using a figure determine whether the following points are
at the right or the left of the line 2x + 3?/  5 = 0: (1, 2), (1,  1),
( 2, 1),(1, 1), (4,  2), (7,  2), (4,  1). Then, draw a figure con
taining the line and the points and verify the results obtained.
2. Find the distance of the point (3, —2) from the line 4x— 3y+6=0.
3. Find the distance of each of the following points from the line
associated with it. In each case interpret the sign of the result.
(a) (2,5),4x + 3y2=0. (e) ( 4, 1), 3^/  2 = 0.
(6) (3, 7),5a; + 12?/+24=0. (/) (a, a), a: + ?/  a = 0.
(c) (2,  2), 3 a:  4 y = 0. {g) (6, a),ax + by = Q.
{d) (5, 2),2x + 5 = 0. (Ji) (1,3), 2/ = 2 a; + 5.
4. Determine the region of the plane defined by each of the following
sets of relations,
(a) a; + 2 ?/ + 4 > 0, (&) 2 x  ?/ + 2 > 0, (c) 2 ic  3 «/ + 6 > 0,
a;_2i/6>0. ?/2<0. 3a; + 2y12<0,
xyl<0.
5. Define by inequalities (as in Ex. 4) the inside of the triangle
■whose sides are given by the expressions in Ex. 4, (c) equated to zero.
6. Define by means of inequalities the inside of the triangle whose
vertices are ( 2, 5), (4, 1), ( 1, 1).
7. Find the distance between the two parallel lines 3x — 62/ + 5 =
and 3x 6?/ 2 = 0.
8. Find the equation of the bisector of the acute angle between the
lines 2a; + 32/4 = 0, x2y + 7 = 0.
314 MATHEMATICAL ANALYSIS [XI, § 203
9. Find the equation of the bisector of the obtuse angle between the
lines in Ex. 8.
10. Prove that the bisectors of the angles formed by the two lines
a\X + 6iy + Ci = and a2X 4 h^y + c^ = are perpendicular to each other.
11. Find the lengths of the altitudes of the triangle whose vertices are
(1,2), (2,3), and (3, 4).
12. Find the area of the triangle in Ex. 11 by multiplying half the
length of one of the sides by the corresponding altitude, and check the
result by finding the area by the formula of § 196.
13. Find the distance of the point (1, 2) from the line 3x + 4 ?/ + 12
= by finding the coordinates of the foot of the perpendicular dropped
from the point on the line and then using the formula for the distance
between two points. Check by means of § 203.
14. If the equations of two parallel lines are ax \hy + c z=Q and
ax 4 by f c' := 0, prove that the distance between them is the absolute
value of (c — c')/y/a^ + h'^.
15. Prove that the bisectors of the angles of a triangle meet in a point.
[Hint : Choose a convenient relation between the triangle and the
axes. ]
16. Find the altitudes of the triangle formed by the lines
x+2?/3 = 0, xy = 0, 4a;yl = 0.
17. Prove that the altitudes on the legs of an isosceles triangle are
equal.
18. Prove that the three altitudes of an equilateral triangle are equal.
19. Prove that the sum of the absolute distances of any point within
an equilateral triangle from the sides of the triangle is constant.
204. Two Equations representing the same Line. If of
two equations of the first degree one can be obtained from the
other by multiplying the latter by a constant, the equations
obviously represent the same line, since all the points which
satisfy one equation must then satisfy the other also. We
now proceed to prove the converse of this statement :
If the equations Ax + By + C = and A'x + B'y { C =
represent the same line, either one can he obtained from the other
by multiplication by a constant.
XI, § 205] THE STRAIGHT LINE 315
Let us suppose first that none of the numbers A, A', B, B\
C, C is zero. The intercepts of the two lines on the icaxis are
then  C/A and  C'/A', on the ySixis  C/B and  C'/B'.
Since the lines are by hypothesis identical, we have
A A' ^. B B'
From these relations follow at once
A _B^_C__.
A'~B~C'~ '
where A; is a constant. It follows that
A^kA', B = kB', C=kC\
If C (or C") is zero, the corresponding line passes through the
origin, and hence the other line must also pass through the
origin ; hence C" (or C) is also zero. We leave the rest of
the proof as an exercise, with the suggestion that the slopes
of the two lines be compared.
205. The Intercept Form. Hesse's Normal Form. We
have called attention thus far to three forms of the equation
of a straight line : (1) the general equation ; (2) the slope
form ; (3) the pointslope form. Two other forms are some
times of great convenience. These are the socalled mtercept
form and normal form. The intercept form is
(11) 1+1='^' ^"^^^'^
where a and h represent, respectively, the x and ^/intercepts
of the line. This equation may be derived by finding the
equation of the line through the points (a, 0) and (0, h). The
derivation is left as an exercise. (See Ex. 21, p. 89.) This
form is not applicable if the straight line passes through the
origin, or if it is parallel to either axis. Why ?
316 MATHEMATICAL ANALYSIS [XI, § 205
The normal form is associated with the name of Hesse,* who
used it extensively. It uses the length p of the perpendicular
droj^ped from the origin upon the line and the angle a which
this perpendicular makes with the a:axis to determine the line.
To derive the equation when p and a are given, we try to
find a relation which is satisfied by the coordiDates (x, y) of
every point P on the line and which is
not satisfied by the coordinates of any
other point. To this end (Fig. 184) we
note that the projection of the broken
line OMP on the perpendicular OQ is
equal to p, if and only if P is on the
line. The projections of the parts OM
and MP on OQ are, respectively, x cos a and y sin a. The
desired equation is, therefore,
(12) X cos a + y sin a = ^
We shall take the positive direction of 0$, or p, from the origin
towards the line, and choose the positive angle XOQ to be a. It is then
evident that the position of any line is determined by a pair of values of
p and a, it being understood that p and a are positive and that a is less
than 360°.
Moreover every line determines a single positive value oi p and a single
positive angle a less than 360", unless p = 0. . When p = {) the line evi
dently passes through the origin and the above rule for the positive
direction of p becomes meaningless. When p — 0, it is customary to
choose a < 180°.
To reduce the general equation Ax { By \ C =0 to the
normal form, we need merely observe that in the latter form
an essential condition is that the coefficients of x and y are
numbers the sum of whose squares is 1, since sin' a  cos^ « = 1.
We must then multiply all the coefficients of Ax {By ^ C =
by a number k, so chosen that (kAy ^(kBy = 1. This condi
♦LuDWiG Otto Hesse (18111874), a noted German mathematician.
XI, §205] THE STRAIGHT LINE 317
tion will be satisfied if ^
Therefore the desired reduction is obtained by dividing the
equation through by ± V^^ + B^, and transposing the constant
term to the righthand side of the equation :
A B ^  (7 _
The sign of the radical must be chosen opposite to the sign of
C, or if (7 = 0, the same as that of B. Why ?
One advantage of the normal form is that every Hne may have its equa
tion written in the normal form. Whether the line passes through the
origin or is parallel to an axis is immaterial.
EXERCISES
1. Reduce the following equations to the normal form. Find in each
case the values of a and p.
(a) 4a; +8?/ 10 = 0. (d) 3x2 y + 6 = 0.
(6) a;!/ + 5=0. (e)y=2x3.
(c) a; + VS ^ = 0. (/) a; = 2y  5.
(gr) The equation of the line whose intercepts are — 5 and 2, respectively.
2. Reduce to the intercept form each of the lines in Ex. 1 for which
such reduction is possible.
3. What are the normal formsof the equations x =3,2 a:f 3 = 0,^—1=0?
4. Derive the process of reducing the equation Ax + By } C = to the
normal form by using the fact (derived from § 203) that p =— Cj^J A^\&.
5. What system of lines is obtained from the normal form, if a has a
fixed value, while p is allowed to assume different values ? If p has a
fixed value and a is allowed to assume different values ?
6. Find the equations of the lines which pass through the point (1, 2)
and are two units distant from the origin.
7. Find the equations of the lines parallel to 5 x + 12 y = 13 and 3 units
distance from it.
8. Find the equations of the lines parallel to 3 x + 4 ?/ = 13 and 7
units distance from it.
318 MATHEMATICAL ANALYSIS [XI, § 205
MISCELLANEOUS EXERCISES
1. Find the equation of the straight line passing through the point
(3, 4), such that the segment of the line between the axes is bisected at
that point.
2. Show that the lines y = ax {■ a, for all values of a, pass through
a fixed point,
3. Given aix + biy + ci = 0, aox + b2y + C2 = 0, agX + bsy + ca = 0,
the equations of three lines forming a triangle. Show that the equation
of any line Ax ^ By ^ C = in the plane may be written in the form
ki{aix + biy + Ci) + koiazx + b2y + C2) + ks^asx + bsy + ^3)= 0,
where ki, k2, ks are constants.
4. Find the ratio in which the line Sy = 6 — x divides the segment
joining the pohits (6, 1) and (— ^>, 2).
5. Find the equation of the line that passes through the point (1, 7)
and makes an angle of 45° with the line x + 2 ?/ = 1.
6. Find the equation of the line that passes through the point (1, 7)
and makes an angle of — 45° with the line x + 2y = 1.
7. Prove analytically that the perpendicular bisectors of the sides of
a triangle meet in a point.
8. Prove analytically that the altitudes of a triangle meet in a point.
9. Prove analytically that the bisectors of the interior angles of a
triangle meet in a point.
10. Prove analytically that the bisectors of two exterior angles of a
triangle and of the third interior angle meet in a point.
11. Theequationsof two sides of a parallelogram are x— 2 i/=l, x+y=S.
Find the equations of the other two sides if one vertex is at (0, — 1).
12. Find the equation of the line passing through the point (1, 1) and
dividing the segment from (— 7, — 2) to (7, — 1) in the ratio 2:6.
13. Two vertices of an equilateral triangle are (1, 1) and (4, 1).
Find the coordinates of the third vertex. There are two solutions.
14. jdind the equation of the line passing through the point (1, 2) and
intersecting the line x \ y = 4^ at a, distance ^VlO from this point.
15. Find the equation of the line through the point (1, 2) which forma
the base of an isosceles triangle with the sides 2x — y = 1, x { y = I.
16. A straight line moves so that the sum of the reciprocals of its
intercepts on the two axes is constant. Show that the line passes through
a fixed point.
XI, § 205] THE STRAIGHT LINE 319
17. If a straight line be such that the sum of the perpendiculars upon
it from any number of fixed points is zero, show that it will pass through
a fixed point.
18. Find the equations of the sides of the square of which two opposite
vertices are (3, — 4) and (1, 1).
19. Derive the formula for the distance of a point (ici, ^i) from the
line^x i By + C = hy finding the intersection of the perpendicular
through the given point and the given line, and then using the formula for
the distance between two points.
20. Prove that if the sum of the first moments of n points with respect
to each of two given perpendicular lines is zero, the sum of the moments
of tliese points with respect to any line in the plane through the inter
section of the given lines is zero. (See Ex. 19, p. 305.)
[Hint : Take the given perpendicular lines to be the axes of
coordinates. ]
21. If with the center of gravity of n points in a plane is associated
the sum of the weights of the n points, prove that the sum of the first
moments of the n points with respect to any line in the plane is equal to
the first moment of the center of gravity with respect to the same line.
22. Given two halflines r, s issuing from a point P, a third halfline t
through P is completely determined if the ratio sin (r^)/sin (ts) = k is
known. The ratio k is called the simple ratio of t with respect to r, s.
Prove that the equations I = and m = of r and s, respectively, may be
so written that, for all positions of t, the equation ot t is I — km = 0.
23. Given two points Pi(xi, y{) and P'z(x2, yi) and a straight line
ax \hy + c = () which meets the line PiPi in Q. Find the simple ratio
[Hint : This can be obtained very readily from a figure by observing
the relation between the desired ratio and the ratio of the distances of
Pi, Pi from the given line.]
24. From the last exercise derive the theorem of Menelaus : If a
straight line cuts the sides of a triangle ABC in three points A\ B', C,
the product of simple ratios
AC BA> CB>
C'B ' A'C' B'A
is — 1. The point A' is on the side opposite A. B' on the side opposite JB,
O on the side opposite C.
CHAPTER XII
THE CIRCLE
206. Review. The circle is the locus of a point which moves
so that its distance from a fixed point, called the center, is con
stant. This constant distance is called the radius of the circle.
If the center of a circle is at the point {h, k) and the radius
is r, the equation of the circle is
(1) {X  hy + (1/  ft)2= r\
Tor, this equation expresses directly the fact that the square of
the distance from the given point (7i, k) to the variable point
{x, y) is r^. Hence, every point on the circle satisfies this
equation and, conversely, any point not on the circle does not
satisfy it.
In particular, if the center is at the origin (h = 0, k = 0), the
equation becomes
(2) x' + y''^ 72^
"We note also that equation (1) when expanded has the form
(3) x''hy^ + Dx + Ey + C = 0.
It follows that every circle in the plane may be represented
by an equation of this form. To what extent is the converse
true? Under what conditions does an equation of the form
(3) represent a circle ? The answer to this question may be
obtained by reference to the method of § 183.
We desire to complete the square on the terms in x, and also on
the terms in y. Therefore we rewrite the equation in the form
ix^ + Dx{ )\.(y'^^Ey+ ) = C.
320
XII, § 207] THE CIRCLE 321
To complete the squares in the two parentheses we need to add
i)Y4 to the first and J5jy4 to the second; to maintain the
validity of the equation we must add the same terms to the
righthand member. We then obtain
or
(:Hf)V(..f)^= + f^.
Since the sum of the squares of two real numbers is positive or
zero, the lefthand member is positive or zero if x, y, D, E are real
numbers. Hence the equation can be satisfied by real coordi
nates X, y only if L^ \ E"^ — ^ C i^ di, positive number or zero.
If Z>2+ ^ — 4 C is positive, equation (3) represents a circle
with center at (— D/2, — E/2) and radius equal to
I y/D^ + E^4.a
If i)2__ ^2 _ 4 (7 ig zero, equation (3) is satisfied by the coor
dinates of the point (— 0/2, — E/2) and by the coordinates of
no other (real) point.
If Z)2 4 ^2 _ 4 (7 is negative, equation (3) represents no real
locus. The answer to our question may then be formulated as
follows : If (3) represents a curve at all, it represents a circle.
207. The Equation of a Circle satisfying given Conditions.
The problem of finding the equation of a circle satisfying
given conditions resolves itself simply into the problem of
determining from the given conditions the values of U, Jc, r in
equation (1), or of D, E, C in equation (3) of § 206. The fol
lowing examples will illustrate the methods that may be used :
Example 1. Find the equation of the circle passing through the three
points (3, — 5), (3, 1), and (4, 0).
The desired equation must be of the form (8), and must be satisfied by
Y
322
MATHEMATICAL ANALYSIS [XII, § 207
the coordinates of each of the three given points. If the first point satis
fies this equation, D, E, and C must be such that
32 +( 5)^ + ■i> • 3 + E(6)+ C=0,
i.e. such that
SD6E+ C=S4.
We find similarly from the second and third of the given points,
3i> + ^+O = 10,
4j[> + C=16.
Solving these three linear equations for Z>, U, C, we obtain
Z)=2, E = 4, C = S.
The desired equation is, therefore,
x2 J ^2 _ 2 ic + 4 2/  8 = 0.
Another method of solving this problem would be to regard (A, A) as
unknown coordinates of the center. They must satisfy the two equations
(3 _ A)2 + (_ 5 _ i.)2 =(3 _ hy +(1 _ j^y2^
(4/i)2+(0 A:)2=(3/02+(l^)2. (Why?)
By solving these equations we can determine h and k. Having found the
center, it is easy to determine the radius. Then the desired equation can
be written down in form (1). The completion of the work here sug
gested is left as an exercise. What other method could be used to solve
this problem ?
Example 2. Find the equation of the
circle inscribed in the triangle ichose sides
are yS=0, 3x— 4?/— 9=0, and 12x+5y
+ 9 = 0.
Let (h, k) be the center of the circle. It
must be equidistant from the three sides.
The distances of (A, k) from the three given
lines are — (A: — 3),  ^ (3 A — 4 A: — 9), and
ji^(12 ft + 5 ^• + 9), the signs being so chosen
that each of these numbers is positive when
(ft, k) is within the triangle. (See Fig. 185.)
By placing the first of these distances equal to tlie second and third, re
spectively, we obtain two equations involving ft and k. The solution of
these two equations yields ft = 1, k — \. Hence the center is the point
(1, 1). The radius is evidently equal to 2. Why? Therefore the
required equation is
(a;  1)2 +(2/  1)2 = 4, or a;2 + y2 _ 2x  2t/  2 = 0.
" X "
Y
^
zzzzzY~.y..iz
HrnTNMTffl
:i=::;z^ii=:=i:
:: zt ::v= : ::
Fig. 185
XII, § 207] THE CIRCLE 323
EXERCISES
1. Write the equations of the circles described below:
(a) Center at the origin, radius equal to 5.
(6) Center at (1, 2), radius = 4.
(c) Center at ( 3, — 2), radius = 3.
(d) Center at (a, a) and radius = a.
(e) Center at (—2, 1) and passing through the point (3, — 2).
(/) Center at (2, 1) and tangent to the xaxis.
2. Discuss fully the locus of each of the following equations :
(a) x^ + y^  2 X + 4:y { I = 0. {d) x'^+ y"^ \ \ =0.
(b) x^ + y^ ix 6y = 0. (e) x^ + y^ \2x 6y + 10 = 0.
(c) x^ \y^ + Sxi = 0. (/) x2 4 2/2 + 2 aa; + 2 a2 = 0.
(g) Sx^ + Sy^\2xiy8=0.
3. What can be said of the coefficients Z>, E, and C in the general
equation if the equation represents a circle which
(a) passes through the origin ?
(6) has its center on the xaxis ? on the yaxis ?
(c) has its center on the line x \y = 0?
(d) touches both axes ?
(e) has its radius equal to 2 ?
4. Find the equations of the circles described below :
(a) Passing through the points (0, 2), (1, 4), (1, 0).
(6) Circumscribing the triangle whose sides are the lines «+ y — 3=0,
x2y + Q = 0, x + 2 = 0.
(c) Inscribed in the triangle whose vertices are (0, 2), (0, — 4), and
(4,1).
(d) Having ( — 2, 4) and (4, — 2) as the extremities of a diameter.
(e) Passing through the points (1, 2) and (2, ]) and having its center
on the line 2x + y + 2 = 0.
(/) Tangent to both coordinate axes and passing through the point
(2, 1). How many solutions are there ?
5. Prove analytically that any angle inscribed in a semicircle is a
right angle.
6. Prove that the locus of a point which moves so that the sum of
the squares of its distances from any number of fixed points is constant
is a circle. Find the coordinates of the center of this circle in terms
of the coordinates of the fixed points. If the number of fixed points is
three, how is the center of the circle related to the triangle whose ver
tices are at the fixed points ?
324 MATHEMATICAL ANALYSIS [XII, § 207.
7. Find the equation of the locus of a point which moves so that the
ratio of its distances from two fixed points is constant and equal to k.
Determine fully this locus. Examine especially the case k=\.
[Hint : Let the two fixed points be (a, 0) and (— a, 0)].
8. Draw the loci of Ex. 7 for different values of k. Prove that if
any one of these loci crosses the line joining the two given points in P and
Q^ respectively, and the raidpoint of the segment joining the given points
is M, we have MP • MQ equal to the square of half the segment.
208. Tangent to a Circle. Point Form. In § 184 we saw
how the slope of the curve Ax^ f Bi/ + Dx \ JEy { C = at
any point {xi, y^ on the curve could be derived. Applying
this method to the circle
x'^j^if^Dx\Ey+ C = 0,
we find the slope m at {x^, y^ on the curve to be
2xy^ + D
2y,\E
The equation of the tangent at the point {xi , y^ is, therefore,
Simplifying, we obtain
(4) 2x]X\2 y^y ^Dx\Ey 2 x^^ — 2y{'— Dx^ — Eyi = 0.
But {xif z/i) is on the curve, and hence
Xi'hyi' + Dx, + Ey,\C=0.
If this identity be multiplied by 2 and added to (4) we obtain
2x,x + 2y,y J^Dx^Ey + Dx, j Ey^ + C = 0,
or
(5) x,x + y,y + 1 D{x + x,)^l E{y + y{)\C=.0,
As a special case of this equation (for D = 0, E = 0, C =
— r^) we obtain the equation of the tangent to the circle
x"^ \ y^= 92 at the point (xi, y^) to be
(6) x,x + y,y = r\
XII, § 209] THE CIRCLE 325
209. Tangent to a Circle. Slope Form. Another form of
the equation of a tangent to the circle x^ \ y"^ = r^ is often very
serviceable. It is derived as follows. The straight line
y = mx \ h meets the circle x"^ \ y"^ = r^ in points whose
abscissas are given by the equation
a;2 I {mx + hf = r».
When expanded this equation becomes
(1 + m2)a;2 ^2mhx \ h"^  r'^ = 0.
The roots of this equation will be real and distinct, real and
coincident, or imaginary, according as
is positive, zero, or negative.
Translated into geometric terms, this means that the line
y = mx + h will meet the circle in two distinct points, two
coincident points, or not at all, according as the expression
above is positive, zero, or negative. If the line meets the
circle in two coincident points, the line is a tangent. The
condition ^ ^^^^  4(1 + m?) {¥  r^) =
yields, after simplification,
or, b = ± rVl + m\
Hence, for all values of m the equation
(7) y = mx ± r VI + m^
represents a tangent to the circle ^ \y'^ — r^.
It follows at once that for all values of m the equation
yk= m(x h)± r VI + m^
represents a tangent to the circle (x — hY\{y — ky = r^.
326 MATHEMATICAL ANALYSIS [XII, § 209
EXERCISES
1. Write the equations of the tangents to the following circles at the
points indicated :
(a) a;2 + ?/2 = 25, at (3, 4).
(&) a;2 + y2 = 6, at(l, 2).
(c) a;2 + ?/2 = 4, at (0, 2).
(d) jc2 + y2 _ 13^ at the points where x = 3.
(e) a;2 4 ?/2 = 10, at the points where y = 1.
(/) a;2 + 2/2 + 2 a:  4 y = 0, at (1, 1).
2. Derive the equation of the tangent to the circle (x — h)^ + (y — k)^
= r2 at the point (xi, yi) by making use of the fact that the tangent is
perpendicular to the radius through the point of contact.
3. Find the intersections of the following circles with the lines in
dicated :
(a) x2y2_ 5andy = 3a;5. (c) ic2+?/2=:i3and3 a:+2 y13=0.
(6) x2 + r/2 = 25 andx  2 !/  5 =0. (c?) x^ + 2/2 = 10 and y = 3 x + 10.
(e) x2 f ?/2 = 4, and y=2x + 4, ?/ = 2x + 2 a/5, y = — 2 x + 5.
Draw a careful figure showing the circle and the three lines.
4. Write the equations of the tangents to the following circles, the
slopes of the tangents being as indicated. Find the points of contact.
(a) x2 + 1/2 = 10, slope =  3. {d) x2 + 2/2 ^ 25, slope = 0.
(6)x2 + 2/2 = 5, slope = ^. (c) (xl)2+(2/+2)2=10,slope=3.
(c) x2 + 2/2 = 13, slope = .
5. Will the equation y = mx ± rVl + m2 represent any tangent to the
circle x2 + 2/2 = r2. Why ?
6. What is the point of contact of the tangent 2/ = wix + rVl + w2
to the circle x2 + y — r^'i From this result derive the equation
a^ia: + y\y = r2.
7. Any circle through the origin has an equation of the form
x^+y^+Dx{Ey=0. Why? Prove that the equation of the tangent at
the origin is Dx^Ey=0. This may be done in at least two different ways.
8. Prove analytically that from an external point two real tangents
can be drawn to a circle.
9. Derive the equation y = mx±r\/l \ m^ directly from the property
that a tangent to a circle is perpendicular to the radius through the point
of contact.
XII, §211] THE CIRCLE 327
210. The Value and Sign of the Expression x^ + y^ + Dx^
+ Eyi + C. The lefthand member of the standard equation
(x — hy +{y — ky = r^ represents the square of the distance
from the point (x, y) to the point (h, k). Hence the expression
(8) (^x^hy+{y,kyr^
is positive, negative, or zero according as (xi, y{) is outside, in
side, or on the circle whose equation is (x — hy \{y — ky = r^.
Moreover, from Fig. 186 it follows that if (xi, 2/1) is a point
outside the circle, the expression (8)
is equal to the square of the length ^ ^{x^,Vj)
of a tangent drawn from the point
(^'d 2/1) to t^® circle. Since the left
hand member of the general equation
x^ \ y"^ \ Dx { Ey \ C = may be writ
ten in the form (x — hy \ {y — ky — r^
we may conclude that the sign of the
^ ^ Fig. 186
expression x^ \ y^ + Dx^ f Ey^ + (7 is
positive or negative according as the point (x^, y^ is outside or
inside the circle x'^{ y"^ \ Dx \ Ey f C= ; and, if positive, it
represents the square of the length of a tangent drawn from the
point (xi, y{) to the circle.
211. The Equations of the Tangents from an External
Point. Suppose we desire to find the equations of the tan
gents drawn from an external point {xi, yi) to the circle
a;2 f 2/2 _ ^2^ Three methods will be discussed:
Example, Find the equations of the tangents drawn from the point
(4,  3) to the circle x^ + y^ = 5.
Method 1. Let {xi, y{) be the point of contact of one of the tangents.
The equation of the tangent at this point is XiX \ yiy = 6 However,
since this tangent passes through the point (4, — 3) we have
(9) 4 xi  3 yi = 6.
328 MATHEMATICAL ANALYSIS [XII, § 211
But the point (cci, yi) is on the circle x^ } y2 _ 5^ Therefore '
(10) xi^ + yi'' = 5.
Solving equations (9) and (10), we find the points of contact to be (2, 1)
and ( — 2/5, — 11/5). Therefore the required tangents are2ic + y— 5 =
and2a; + lly + 25 = 0.
Method 2. From § 209 it follows that any tangent (not parallel to
the ?/axis) to the circle x^ \ y^ = 6 is of the form y = mx ± VsVl + m^.
Since this tangent is to pass through the point (4, — 3) we have
— 3 = 4 m ± VoV 1 j W'^5
which simplifies to 11 m + 24 m + 4 = ; th is giv es m =— 2, or — 2/11.
Substituting these values in y — mx ± VSVl + m^ and simplifying we
have 2x + ?/5 = and 2 x + 11 y + 25 = 0.
Method 3. The equation of any line through the point (4, — 3) is of
the form ?/ + 3 = m(x — 4). Eliminating y between this equation and
x2 + y2 — 5 ^e have
(11) (wi2 + l)x2+ x(8m26m) + (16m2 + 24m + 4)=0.
Now since we desire y + 3 = w(:k — 4) to be tangent, equation (11) must
have equal roots, i.e. ( 8 m2  6 m)2 _ 4(m2 + 1) (16 m2 + 24 m + 4) =
or 11 m2 + 24 m + 4 = which gives m=— 2, or —2/11. Therefore
the equations of the tangents are 2x + ?/ — 5 = and 2x+lly+25 = 0.
212. The Polar of a Point with respect to a Circle. Let
us apply the first method of § 211 for finding the equations of
the tangents from an external point to a circle, to the general
problem of finding the equations of the tangent from the point
(xi, 2/1) to the circle x^ \ if = r^. The coordinates {x\ y') of the
point of contact are then found by solving simultaneously the
pair of equations x'x^ + y'yi = r^, x'^ f y'^ = r\ The first equa
tion expresses the fact that the point (xi, y^ is on the tangent
x^x { y'y = r"^ ', the second, that {x\ 2/') is on the circle.
This shows that the straight line XiX + yiy = r^, where (xi, y{)
is any external point, meets the circle in the points of contact
of the tangents drawn from {xi, y^. In other words,
(12) iCiic + ;viy = 72
is the equation of the line joining the points of contact of the
XII, § 212]
THE CIRCLE
329
tangents through (xi, 2/1), if the latter point is outside the circle.
If this point is on the circle, we know that (12) is the equation
of the tangent at the given point. Finally, if (xi, y^ is inside
the circle, (12) represents a definite straight line determined
by the point and the circle. This straight line (12), whether
(^1) 2/1) is outside, on, or inside the circle, is called the polar of
Fig. 187
(s^ij 2/1) with respect to the circle. The polar of (a^i, y^ with
respect to a circle is then a uniquely determined line for every
point (aji, 2/1) in the plane, except the center of the circle.
(See Fig. 187.) Why this exception ?
EXERCISES
1. Are the following points inside, outside, or on the circle x^ + y^
2x + 6y15=.0? (1,2), (1,0), (1,4), (3,0), (3,0), (0,2),
(5, 1). For the points outside, find the length of the tangents drawn to
the circle. Draw carefully a figure to illustrate each of your results.
2. What is the length of the tangents drawn from (1, 1) to the circle
whose equation m 2 r^ \ '2, y'^ { ^ x — b y — \ = Q?
[Caution : The equation is not in the standard form.]
3. Find the equations of the tangents drawn from the following points
to the circle indicated :
(a) ( 2, 4) ; a;2 + y2 = 10. (d) (3, 2) • x'^ + y^ = 4.
(6) (5, _ 1) ; x2 + y2 = 13. (e) (4, 3) ; x2 + y2 = 16.
(c) (3,  1) ; x2 + 2,2= 2. (/) (7, 1) ; x2 + 2,2 ,, 25.
4. Find the equations of the tangents drawn from (0, 4) to the circle
aj2 + y2 _ 2a; + 6y15 = 0.
6. Show that the polar of a point P with respect to a circle is per
pendicular to the radius or radius extended through the point P.
330 MATHEMATICAL ANALYSIS [XII, § 212
6. Show that if P is inside the circle, the polar of P is wholly outside
the circle.
7. Show that if the polar of P with respect to a circle whose center is
cuts the line OP in Q, then OP • OQ=r^, where r is the radius of the circle.
[Hint : Let the center O be the origin and the line OP the xaxis.]
8. Show that if the polar of a point with respect to a given circle is
given, the point is uniquely determined.
[Hint : This follows directly from the results of Exs. 5 and 7 ; or it
may be proved directly by identifying the given polar ax\by + c =
with the equation Xix + y\y = r^. In the latter case we should have
Xi/a = pi/b = — r^/c, which determines xi, yi uniquely.]
9. A straight line is drawn through a given point P, cutting a given
circle in the points A and B. Calculate the length of the segments PA
and PB. Let P be chosen as origin and the line through P and the
center of the circle as icaxis. The equation of the circle is then x^ + y'^
4 Dx 4 C = 0. If p is one of the segments PA or PB and « is the
angle which PA makes with the xaxis, the coordinates of J. or P are
{p cos a, p sin a). Since this point is on the circle we have the equation
(/) cos a)2 + {p sin ay + !>/) cos « + O =
for determining the two values of p. This equation reduces to
p2 f Z) cos a . /) + = 0.
It may be noted that the product of the roots pip^ of this equation is C,
i.e. independent of a. What theorem of elementary geometry does this"
prove ? Prove also that the product PA • PB is positive or negative
according as P is outside or inside the circle.
213. The Intersection of Two Circles. Given two circles
0^2 f 2/2 4 D.a: + ^i2/ 4 Ci = 0,
and ic2 4 2/2 + D.p: + E^ + 0^= 0.
The coordinates of the points of intersection are found by
solving the equations simultaneously. Subtracting the equa
tions, we have
(AA)a5+(^i^2)2/ + c,a = o.
Every point common to the two circles will satisfy this last
equation, which is the equation of a straight line. Therefore
the problem of finding the points of intersection of two circles
XII, § 214] THE CIRCLE 331
is equivalent algebraically to that of finding the intersections of
a straight line and a circle. This problem leads essentially to
the solution of a quadratic equation in one unknown. There
fore we may conclude that two circles may intersect in two
distinct points (two real roots), may be tangent to each other
(coincident roots), or may not intersect at all (imaginary roots).*
214. Orthogonal Circles. Two circles which intersect at
right angles are said to be orthogonal. In this case the tan
gents to the two circles at a point of
intersection must be perpendicular, and
the two tangents pass respectively through
the centers of the circles (Fig. 188). The
condition for orthogonality is then simply
that the sum of the squares of the radii _ .„^
Fig. 188
of the circles shall be equal to the square
of the distance between their centers. If the centers are
(7i(7ii, ^i) and €2(712, k^ and the radii are Vi and r^ respectively,
the condition for orthogonality is
If the equations of the circles are
,. ox ^' + y' + ^^^ + ^^y + c'l = o»
^ ^ x^ + y'^ + D,x + Eiy+C2 = 0,
this condition becomes (see § 206)
4 4 4 "^ 4 '
which when simplified gives
A A + A^2  2(Ci + C2) = 0.
* The reasoning above breaks down, if Z)i  2)2 = and Ei — E2 = 0, that is
when the circles are concentric. In this case, unless C\ —0^ = also (in which
case the two circles coincide), the two equations are inconsistent and have no
common solution, real or imaginary.
332 MATHEMATICAL ANALYSIS [XII, § 215
215. Pencil of Circles. Let the lefthand members of the
equations (13), § 214, be represented by Mi and Mz respectively.
Let us consider the locus of the equation
(14) MikM2 = 0,
where k is an arbitrary constant. This equation may, iik^l,
be written in the form
(16) x^ + 2,^ + L3^^» + L_^^2/+Yr = 0.
which represents a circle for each value of k{=^l). When
k = l, equation (14) represents the straight line
(16) (Z>i  D,)x + {El  E,)y + Ci  C^ = 0.
The system of circles obtained by giving different values to
A;, is called the pencil of circles determined by the two given
circles. The straight line (16) is called the radical axis of
the two given circles, and of the pencil.
The following properties of a pencil of circles are readily
proved :
If the tivo gwen circles intersect in two points A and B, every
circle of the pencil passes through A and B.
If the two given circles are tangent to each other at a point Ay
all the circles of the pencil are tangent at A.
Tf trough any point in the plane not on the radical axis of the
circles passes one and only one circle of the pencil. The proofs of
these theorems are left as exercises.
Further properties of pencils of circles will be found in the
following exercises.
EXERCISES
1. Find the coordinates of the points of intersection of the following
pairs of circles :
(o) x2 + y2 ^ 5 and a;2 + y2 ^ 2 X  4 2/ + 1 = 0.
(6) x2 + y2 _ a; _. 2 J/ = and x^ + y2 4. 2 x — 4 y = 0.
(c) x2 + y2 4. 2 X  17 = and x2 + t/2 _ 13 _ q.
XII, § 215] THE CIRCLE 333
2. Write the equation of the radical axis of each pair of circles given
in Ex. 1.
3. Prove that the tangents drawn from any point of the radical axis
of two circles to the two circles are equal.
4. Prove that the circles x^{y'^\6x — 2y + 2 = and x'^ +y'^ + 4y
+ 2 = are tangent to each other. Find their point of contact and the
equation of their common tangent.
5. Find the equation of the circle through the intersections of the
circles x^+y^ — 4x — 4 = and x^ + y^+2x — Qy — 2 = and the point
(3, 3). [It is not necessary to find the intersections.]
6. Prove that the following circles are orthogonal: x^\y^ — 2x—4=0
and x^+y'^—Q ?/+4=0. In general for the circles : x^ \ y'^ \ Dx — C =
and x^ + y"^ + Ey + C = 0.
7. Determine C so that x^ + y^ — 2x + 4:y — S = and x^ \y^ \2x
+ (7 = are orthogonal.
8. Prove that the locus of the centers of the circles of a pencil is a
straight line perpendicular to the radical axis of the pencil.
9. Prove that if the radical axis of a pencil of circles is chosen as the
yaxis and the line of centers as the icaxis, the equation of any circle of
the pencil is of the form x^ + y'^ + kx \ C = 0, where C is the same for all
circles of the pencil ; and that all circles obtained by varying k in this
equation are circles of the same pencil.
10. The circles of the pencil in Ex. 9 intersect in distinct points, are
tangent to each other, or do not intersect at all, according as C is negative,
zero, or positive. In case (7 = 0, all the circles of the pencil are tangent
to one another at the origin. Draw carefully three figures, illustrating
the three kinds kinds of pencils here indicated.
11. Find the equation of a circle wbich is orthogonal to two given
circles of the pencil in Ex. 9.
[Hint : Let the two given circles be
x"^ + y^ + kix + C=0 and x^ +y^ + kix + C = 0,
and let the required circle be x"^ + y'^ + D^x + E^y + (^2 = 0. If this circle
is to be orthogonal to each of the given circles we must have (§ 214)
B^kx  2((7 + (72) = and D^iki  2(0 + (^2) = 0.
These equations give 2)2 = and d^C. Hence the required equation
is x2+y2__^2?/— C=0. This yields two remarkable results : (1) The coeflB
cient E2 is undetermined, and by varying E^ we have a pencil of circles
each of which satisfies the condition of being orthogonal to the two given
334
MATHEMATICAL ANALYSIS [XII, § 215
circles. (2) The equation found is independent of An, and k2 Hence,
every circle of the pencil just found is orthogonal to each of the circles of
the given pencil. Writing I for E2 to obtain uniformity of notation, wre
have found two pencils of circles :
and
x2 + 2/2 4 A:x + O =
«2 + 2/2 + ;y _ o = 0,
such that every circle of either pencil is orthogonal to each circle of the
other pencil. These two pencils of circles are said to form an orthogonal
system. (See the adjacent figure.)]
12. In an orthogonal system of circles, the centers of the circles of one
pencil are on the radical axis of the other pencil.
13. If the circles of one pencil of an orthogonal system intersect in two
distinct points A and J5, the circles of the other system do not intersect at
all, but pass between the points A and B.
14. If the circles of one pencil of an orthogonal system are mutually
tangent to each other at a point A, the circles of the other pencil are also
mutually tangent at A.
15. Prove that the three radical axes of three circles (not belonging to
the same pencil) taken two by two intersect in a point. This point is
called the radical center. Show that it is the center of a circle orthogonal
to each of the three given circles and that the tangents drawn from it to
the given circles are equal.
XII, § 215] THE CIRCLE 335
MISCELLANEOUS EXERCISES
1. Find the condition that ax { by \ c = be tangent to the circle
«2 + y2 _ r2.
2. Find the equation of the circle passing through the points (0, 0),
(a, 0), and (0, &).
3. Show that the equation of the circle having the points (a^i, yi) and
(^2, 2^2) as the extremities of a diameter is (x— Xi)(x — a^a) + (.V — Vi)
(y  2/2) = 0.
[Hint : The circle is the locus of the vertex of a right angle whose
sides pass through the given points. ]
4. Find the equation of a circle w^hich is tangent to the lines a; = 0,
y = 0, and ax + by \ c = 0.
5. A line is drav^^n through each of the points (a, 0) and (—a, 0),
the two lines forming a constant angle d. Find the equation of the
locus of their point of intersection.
6. A straight line moves so that the sum of the perpendiculars drawn
to it from two fixed points is constant. Show that it is always tangent to
a fixed circle.
7. Give a geometrical construction for the polar of a point with
respect to a circle.
8. If the polar of a point P passes through Q, then the polar of Q
passes through P.
9. Find the equations of the common tangents of the circles x^+y^=S
and x2 + ?/2 _ 10 ic + 20 = 0.
10. Find the locus of a point which moves so that the length of a tan
gent drawn from it to one given circle is k times the length of a tangent
drawn from it to another given circle.
11. Find the equation of a circle through the points of intersection of
3:2 +2/2 _ 4 and x^\y^—2x\4:y+4:=0 and tangent to the line x—2y=0.
12. Show that the polars of a given point P with respect to the circles
of a pencil pass through a fixed point, unless P is on the line of centers.
13. A point moves so that the sum of the squares of its distances from
the sides of a given square is constant. Show that its locus is a circle.
14. A point P moves so that its distance from a fixed point A is always
equal to k times its distance from another fixed point B. Show that its
locus is a circle, if k =^1. Show also that for different values of k
these circles have a common radical axis.
336 MATHEMATICAL ANALYSIS [XII, § 215
16. A line rotating about a fixed point meets a fixed line in a point
P. Find the locus of a point Q on OP such that OP • OQ is constant.
16. Prove that among the circles of a pencil there are at most two which
are tangent to a given straight line (unless all the circles are tangent to
the line) . When is there only one ? None ?
[Hint : Let the given line be the scaxis.]
17. Inversion with Respect to a Circle. Given a circle with center
and radius r. Corresponding to any point P in the plane (distinct from 0)
there exists a unique point P' on OP such that OP • OP' = r^. The
point P' is called the inverse of P with respect to the given circle. Prove
the following propositions :
(a) If P' is the inverse of P, P is the inverse of P'.
(b) If P is inside the given circle, P' is outside ; and vice versa.
(c) Every point on the given circle corresponds to itself.
(d) If the coordinates of P and its inverse P' are (x, y) and (x', y')
respectively, referred to two rectangular axes through 0, we have
x'=J^, y' = '^; and x= ^^^' , y= ^^^^ •
(e) If a point P describes a curve, the inverse P' describes a curve
called the inverse of the former curve. The inverse of any straight line
through is this line itself.
(/ ) The inverse of any line not through is a circle through O, and
the inverses of parallel lines are circles tangent at 0.
(g) The inverse of any circle is a circle, unless the given circle passes
through 0, in which case its inverse is a straight line.
(h) Two orthogonal circles or lines have orthogonal inverses,
(i) Any circle orthogonal to the given circle is its own inverse.
(j) The adjoining figure illustrates a
simple mechanism for changing circular
motion into rectilinear motion. It is known
as the inversor of Peaucellier. The heavy
lines represent rigid bars, hinged at their
extremities. The sides of the quadrilateral
ABCD are all equal and OB  0D= p.
Prove that if O.is fixed and the mechanism
is allowed to move in any way it can, C is
always the inverse of A with respect to a circle with center O and radius
r = \JPp^, where I is the side of the rhombus ABCD. Hence, if A de
scribes a circle through 0, G will describe a straight line.
CHAPTER XIII
THE CONIC SECTIONS
216. Definition of a Conic. A conic section* or simply a
conic is defined as the locus of a point which moves so that its
distance from a fixed point, Fi, is always equal to a given
constant, e, times its distance from a fixed line D^D^^.
The fixed point F^ is called the focus. The fixed straight
line A A' is called the directrix. The constant e is called the
eccentricity. It is assumed that e > and that F^ does not
lie on A A'.
If P (Fig. 189) is any point on the curve, we have, by
the preceding definition,
(1) F,P = e' MP,
where MP is the perpendicular distance of P from the
directrix. It must be remembered that F^P and MP are
absolute quantities, not directed quantities, and that e is
positive.
♦ The name "conic section " is due to the fact that the curves in question
were originally obtained as the sections of a right circular cone. They
were discussed from this point of view by the ancient Greeks.
z 337
338 MATHEMATICAL ANALYSIS [XUl, § 217
217. The Equation of a Conic. Let the directrix be chosen
as the 2/axis and the line through Fi perpendicular to the
Y directrix as the ajaxis (Fig. 190). The coordi
nates of Fi may then be taken as (), 0),
y *' where p is different from zero. Let P (a;, y)
be any point on the conic. Then
A'
P F^P.o)X
F,P = ^{xpy + y%
Fig. 190 ^^^ ^^ = + a; or  a;
according as x is positive or negative. Equation (1), § 216
then becomes
V(a;i))2 + 2/2 = ±ea;.
Squaring both sides of this equation and simplifying, we have
(2) (1  e2)a;2 + 3/^  2px{p'^ = 0.
This is the equation of the conic. For, the coordinates of
every point {x, y) satisfying the definition of the conic will
satisfy equation (2), and conversely, every point whose coor
dinates satisfy equation (2) will satisfy equation (1). Why ?
This is an equation of the type considered in § 183. It
represents an ellipse if 1 — e^ > 0, a hyperbola if 1 — e' < 0,
and a parabola if 1 — e'^ = 0. Hence we have,
A conic is an ellipse, a parabola, or a hyperbola according as
the eccentricity e is less than 1, equal to 1, or greater than 1.
THE ELLIPSE
218. Standard Equation of the Ellipse : e < 1. We have
seen in § 183 how to determine the locus of equation (2)
by completing the square. If we apply the same method
here, equation (2) may be written in the form
(3)
(1e^
XIII, § 218]
THE ELLIPSE
or
(4)
X —
P T , 2/' _
pH^
ri _ p2\ ' 1 _ p2
^1 — ^2\2
339
Since 1 — e^ is positive by hypothesis this equation represents
an ellipse whose center is at the point {p/{l — e^), 0), and
whose axes coincide with the two y\
straight lines x = p/(l — e^) and y =
(Fig. 191).
Let us move the curve parallel
to the a;axis through a distance
p/(l  e2), i.e. to the left if p > 0.
Then its center comes to the origin,
and its equation becomes
Fig. 191
(5)
or
(6)
x^ +
p^e
2p2
1e^
y2
(1  e^y
p^e^
p2^2
= 1.
If we place
(1  e2)2 1
(7)
(162)2
= «,
j9%2
= 6S
the equation of the ellipse in its new position, i.e. with its
center at the origin (Fig. 192), becomes
Fig. 192
(I.)
a2 &2
From (7) we have
(8) 62=^2(162),
which shows that b <a, since e < 1.
If the ellipse is given in the form
(IJ, a and b are known. Then the
340 MATHEMATICAL ANALYSIS [XIII, § 218
value of e can be found in terms of a and b bysolving equation
(8) ; this gives
(9) e^ = ^^A^
219. Properties of the Ellipse. It is important to distin
guish between the properties of a curve as such and those
properties which are concerned merely with the relations the
curve bears to the coordinate axes. Thus the ellipse, as a
certain kind of curve, is symmetrical with respect to two
perpendicular lines called the axes of the curve. The longer
of the segments on these lines cut off by the curve is called
the majon axis, the shorter one, the minor axis. The inter
section of the two axes of the curve is called the center of
the ellipse.
Every ellipse, no matter how it is situated in the plane
of coordinates, has a major axis and a minor axis as well as a
center. From the way in which the equation was derived, we
know also that every ellipse has a focus and a directrix. The
symmetry of the curve with respect to the y'dxis shows that
this same curve could have been obtained from a second focus
F2 and a second directrix D2D2 on the opposite side of the
center.
We shall now investigate how the two foci and the two
directrices are related to the major axis, the minor axis, and
the center.
220. Foci and Directrices. The original position of the
focus Fi was (p, 0) ; the abscissa of its new position is
^ 1 _ e2 1 _ e2
Since from (7) we know that pe/{l — e^) = a, we find the
coordinates of the focus F^ in the new position to be ( — ae, 0).
XIII, § 221]
THE ELLIPSE
341
(See Fig. 193.) Similarly the equation of the directrix AA'
in its new position is
X =
P
or
(10)
__a
e
The second focus F2 has the
coordinates (ae, 0). The second
directrix D2D2 has the equation
FiQ. 193
(10')
a
X=~'
e
221. The Ellipse in Other Positions. If the center of
the ellipse is at the origin and the major axis is on the yaxis,
the equation of the ellipse is
(I.)
62 ^ a2 '
where, as before, 2 a is the length of the major axis and 2 6 is
the length of the minor axis. The foci of this curve are at the
points (0, ae), (0, — ae) ; the equations of the directrices are
2/ = ± « A
The equation of an ellipse whose center is at the point
(h, k) and whose axes are parallel to the coodinate axes is
(II.) (£^+(1^=1, (a>6)
or
(II„) .(l^ + (L=^^i, (a>b)
according as the major axis is parallel to the a>axis or to the
ySixis. Finally we can reduce an equation of the form
(III) Ax^hBy^ + Dx^Ey+C^^O, A>0,B>0,
to the form II_, or 11^^, if it has a real locus. (See § 183.)
342
MATHEMATICAL ANALYSIS [XIII, § 222
222. The Case a = b. The Circle. If a = b the equation
(Ij.) reduces to the equation of a circle. The relation a=b
implies, however, that e = and this value of e is excluded in
the definition of a conic. On the other hand it is clear that
for a given value of a, as the eccentricity approaches zero, the
ellipse approaches a circle. At the same time, the foci ap
proach the center, and the directrices recede indefinitely.
Why ? Since the circle is a limiting form of an ellipse it is
classified as an ellipse with equal axes and is counted among
the conies.
223. A Geometric Property of an Ellipse. An important
geometric property of any ellipse follows from the fact that
the distance from the center to either focus, which we shall
denote by c, is given by the relation
or
(11)
c=:ae= V a^
^S
This relation shows that c, a, and h are the sides of a right
angled triangle in which a is the hypotenuse (Fig. 194). In
other words, a circle drawn with its center
at an extremity of the minor axis and with
its 7'adius equal to a, will cut the major axis
in the foci, Fi and F2.
In computing the elements of an
ellipse from a and b, it is generally con
venient first to find c from (11) and then
to find e from the relation*
(12)
e = £
* This relation is equivalent to (9), § 218. It may be expressed by saying
that e is the cosine of the angle CF^B, Fig. 194.
XIII, § 224]
THE ELLIPSE
343
The extremities of the major axis are called the vertices of
the ellipse.
The chord through a focus perpendicular to the major axis
is called a latus rectum. Its length is 2 b^/a. Why ?
_
^ ^ ■■'^=
W'' _ ^ s
_, 5^^
I ^ Z .5
•3,22 * ^ 2 no)
. _C:fiQ) . J^Q) JC
S (^
^^=, =^
rZ
i ^
±_ » :
Fig. 195
224. Illustrative Examples. Example l. Given the ellipse
4a;2 + 9y2_36 = 0.
Find the coordinates of the center, the vertices, the foci, and the equa
tions^ of the directrices.
The given equation may be written in
the form
9 4
from which follows that a = 3, 6 = 2.
Therefore c = Va^ — b^ = \/5 and e = V5/3.
The coordinates of the center are (0, 0),
the vertices (3, 0) and (—3, 0), the foci
(— V5, 0) and ( VS, 0) and the equations of
the directrices are x = — 9/ V5 and x = 9/ VS
(Fig. 195).
Example 2. Find the coordinates of the center, the vertices, the foci,
and the equations of the directrices of the ellipse
25 x* + 9 2/2  50 x+S6y  164 = 0.
From § 183, we know that the given equa
tion may be written in the form
25(x  1)2 + 9(?/f 2)2=225,
or
(x1)^ , (y+2)2 ^.^
9 25
We now conclude that the center is at
(1, — 2), and that the major axis is parallel
to the yaxis. Here a = 5, 6=3, c = 4, e = 
Fig. 196 and a/e = ^^. Sketching the ellipse we find
from the figure that the vertices are (1, 3)
— 7), and the foci (1, 2) and (1, —6). The equations of the
33/4.
^ , "
~^4
'&"
^^ ^s^
^ 0.2} ^
t X
^"
(/r2)
A ^ i
\ "^^
^^^a:
._ : ^± :
::::i:i±li^:
and (1,
directrices are y = 17/4, y
344 MATHEMATICAL ANALYSIS [XIII, § 224
EXERCISES
In the following ellipses determine the major axis, the minor axis, the
coordinates of the center, the coordinates of the vertices and foci, and the
equations of the directrices. Sketch the curves.
1. 3 a2 + 4 ?/2 = 12. 7. 3 a;2 + 3 2/2 = 12.
2. 4 a:2 + 3 ?/2 = 12. 8. x^ + 2y^= 8.
3. 4x2 + 2/2=16. 9. 4x2 + 9 2/216a;18?/23=0.
4. 36 x2 + 25 2/2 = 144. 10. 9 a;2 + 25 2/2  150 y = 0.
5. 2 x2 + 4 2/2 = 3. ^ 11. 4 a;2 + 2/^  8 a; + 4 2/ + 4 = 0.
6. 5x2 + 2/^^=75. 12. 9x2 + 42/2 + 36x16 2/+16=0.
13. Write the equation of the following ellipses :
(a) Center at origin, major axis = 4 on xaxis, minor axis = 3.
(&) Center at origin, major axis = 5 on 2/axis, minor axis = 3.
(c) Center at origin, major axis = 6, minor axis = 3 (two solutions).
(d) Center at origin, eccentricity 4/5, foci at (—2, 0) and (2, 0).
(e) Center at (1, 2), major axis = 6 parallel to xaxis, minor axis = 4.
(/) Foci at (0, 2) and (0, 8), major axis = 10.
14. An ellipse has its center at the origin, and its axes coincide with
the coordinate axes. The ellipse passes through the points ( VT, 0) and
(2, 1). Find its equation.
[Hint. Assume the equation of the ellipse in the form (1^.). Find a
and b from the fact that the ellipse must pass through the given points.]
15. Find the equation of the ellipse symmetrical with respect to the
coordinate axes if the major axis is twice the minor axis and the curve
passes through the point (2, 1). How many solutions ?
16. Show that the equation of the ellipse whose vertex is at the origin
and whose major axis is on the xaxis is of the form a'^y = ?>2(2 ax — :e2).
17. Verify equation (I^) by deriving the equation of a conic whose
focus is at (— ae, 0) and whose directrix is the line x =— a/e.
18. Find the equation of the ellipse whose focus is at (0, 0), whose
directrix is the line x + 2/ — 1 = and whose eccentricity is 1/2.
19. Find the equation of the ellipse whose eccentricity is 1/3, whose
focus is at (3, 1) and whose directrix is the line 3x + 42/ — 1=0.
20. Find the equation of the conic whose focus is at (2, 1), whose
eccentricity is 3, and whose directrix is the line 3 x + 2/ = 1. What kind
of a conic is the curve ?
XIII, § 225]
THE ELLIPSE
345
225. Focal Radii. The segments F^P and F2P joining any
point P on an ellipse to the foci JF\, F.^, are called the focal
radii of the point P.
If the equation of the ellipse is given in the standard form (/,),
the focal radii of any poi7it P{xi, 2/1) «^e a — ea'i, a + exi.
A
Y
P (x„v,\
D,
M,
M,
r>
/K
4
\^i
F.J
X
Fig. 197
For, from the definition of an ellipse (Fig. 197),
• FiP=e'M,P, F.P=e'PM.^
But from the figure, we have also
M,P='^ + x,,
PM2 = Xi.
e
Therefore the focal radii are
FiP = a f ex„ F2P —a — exi.
From these relations follows the important property :
The sum of the focal radii of any point of an ellipse is constant
and is equal to the major axis 2 a.
It may be noted that this relation still holds when the
ellipse is a circle (e = 0), although the method of its derivation
is not applicable in this case. An ellipse could, therefore, be
defined as the locus of a point which moves so that the sum of its
distances from two fixed points (the foci) is constant.
346 MATHEMATICAL ANALYSIS [XIII, § 226
226. Geometric Constructions of the Ellipse. The property
of the ellipse derived in § 225 gives the construction indicated
in Fig. 198 for the points of the ellipse when the foci and
the major axis are given.
1 ^~r^
Fig. 198
The segment AB is the major axis. Different positions of P
on this segment give corresponding values AP and PB of the
focal radii of a point on the ellipse. Circles drawn with these
radii and centers at the foci intersect in points of the ellipse.
To each position of P on AB correspond four points of the
ellipse.
A very convenient method of drawing an ellipse is indicated
in Fig. 199. Two pins are stuck in the paper at the foci and
Fia. 199
a loop of thread thrown over them. If a pencil point is in
serted in the loop and moved so as to keep the thread taut, it
will describe an ellipse. Why ?
Another method of constructing an ellipse (much used by
draftsmen) is based on the fact (§ 179) that if the ordinates of
the circle x"^ + 2/2 = a^ are shortened in the ratio b : a (b < a)
XIII, § 226] THE ELLIPSE 347
there results an ellipse with major axis 2 a and minor axis 2 b.
The adjoining figure (Fig. 200) exhibits the method. Explain
and prove the method correct.
Fig. 200
[Hint. The two circles being of radii b and a respectively, we have
OB/OQ = b/a ; hence, MP/MQ = b/a. Why ?J
EXERCISES
1. Construct an ellipse whose foci are 2 inches apart and whose major
axis measures 3 inches.
2. Construct an ellipse whose major and minor axes are 2 and 1.6
inches respectively.
3. From the property of § 225 derive the equation of an ellipse.
4. From Fig. 200 show that the coordinates (x, y) of any point on the
ellipse (Ij.), p. 339, are given bj'^ the equations
x= acosQ, y = b sin 0,
where 6 is the angle MOQ. Do these values of x, y satisfy the equation
of the ellipse for all values of ^ ?
5. From the relation between the ordinates of a circle and an ellipse
whose major axis is equal to the diameter of the circle prove that any
plane section of a circular cylinder is an ellipse, provided the plane of
section is not parallel to an element of the cylinder.
6. Prove from the result of the last exercise that a properly determined
plane section of an elliptic cylinder is a circle.
348
MATHEMATICAL ANALYSIS [XIII, § 227
THE HYPERBOLA
227. Standard Equations of the Hyperbola. If e > 1, then
1 _ e2 < 0, and it is convenient to write (2), § 217, in the form
(13)
{f — V)x^ — 'f\2jixf'^ 0.
Completing the square and transforming as in § 218, we
obtain / \2 2 ^^ra.
This equation represents a hyperbola whose center is at the
point ( — i)/(e^ — 1), 0) and whose axes coincide with the lines
x =  j>l{f  1), and 7/ = (Fig. 201).
\:.
a
Xi
\ 6
a >y
A
^n
y.
/"
K
\v^\(fl^^o)*x
/\
•>[
X
Fig. 201
Fig. 202
If the curve is moved parallel to the a>axis so that its center
coincides with the origin (Fig. 202), its equation becomes
y
^2g2
p^e^
= 1.
(e2_i)2 e2_i
If, then, we place
the equation of the hyperbola becomes
(Ix)
XI II, § 2271 THE HYPERBOLA 349
' Erom (14) we have the relation connecting a, b, e as
62 = a2(e2l),
or
(15) e^ = ?i±*?.
Here, as in the case of the ellipse, it is important to note
some of the properties of the curve. It is seen that the
locus is symmetrical with respect to the line passing through
the focus and perpendicular to the directrix. This line is called
the principal axis and the segment of this line intercepted by
the curve is called the transverse axis and its length is 2 a.
The extremities of the transverse axis are called the vertices,
and the point midway between the vertices is called the center.
The curve is also symmetrical with respect to the line through
the center and perpendicular to the transverse axis. The seg
ment on this line whose length is 2 6 and whose midpoint is
at the center of the hyperbola is called the conjugate axis.
If a hyperbola has its center at the origin, and if its trans
verse axis 2 a is on the ?/axis, and its conjugate axis is 2 b, its
equation is
(I) ^y=i.
The equation of a hyperbola whose center is at the point
(h, k), whose transverse axis is 2 a, and whose conjugate axis
is 2 b, is
(IIJ (^_(J^ = 1, or (II.) (l^(l^^ = l,
according as the transverse axis is parallel to the xaxis or the
yaxis.
The equation of any hyperbola with axes parallel to the
coordinate axes may be written in the form
(III) Ax^\By^'hDx + Ey\C=X), A>0,B<0;
350
MATHEMATICAL ANALYSIS [XIII, § 227
and every equation of this form (A > 0, B < 0) represents a
hyperbola or a pair of straight lines (cf. § 183).
As in the case of the ellipse, it is easy to show that every
hyperbola has two foci on the transverse axis, one on each
side of the center and at a distance
c from the center, where
c2 = 02^2 = a2 I bK
With each focus is associated a
directrix perpendicular to the trans
verse axis and at a distance a/e
from the center (Fig. 202).
The latus rectum, i.e. the chord
through the focus and perpendicular to the transverse axis pro
longed, is of length 2 b^/a. The asymptotes of the hyperbola
Fig. 202 (repeated)
(xhy {yky ^
¥
= 1
are the lines
(16)
{^hY
&2
228. Geometric Properties of the Hyperbola. Tlie segment
from the center to a focus of a hyperbola is the hypotenuse of
a rightangled triangle ivhose legs are the semitransverse and
seniicoiij agate axes. Why? It is readily seen, moreover,
that, if a rectangle be constructed by drawing lines through
the extremities of each axis parallel to the other axis, the
diagonals (extended) of this rectangle are the asymptotes of
the hyperbola (Fig. 202). The circle drawn on either diagonal
as a diameter passes through the foci. Why ?
229. Illustrative Examples.
Example 1. Find the coordinates of the center, the vertices, and the
foci, and the equations of the directrices and the asymptotes of the hyperbola
4x29.^2 + 36 = 0.
XIII, § 229]
THE HYPERBOLA
351
The equation is readily transformed into the form
9 4
It is now seen that the center is at the
origin and that the transverse axis is along
the yaxis (Fig. 203). The vertices are
(0,2) and (0, 2). Since c = V^, the
coordinates of the foci are (0, Vl3) and
(0, — Vi3). The asymptotes are given
by 4a;2_9y2_o or 2x Sy = and
2x\Sy = 0. Since e = \/l3/2 the equa
tions of the directrices are
y = ±
Vis
±±vrs.
ss^ F. (h, ^) >j/
i j^X'Z^t M
s / T
^o^^ 1 ?
D^ ^i=^r ?i
^i^^ ^2^^N
<^^ ^'^ ^^^^
y^ tTa^tS^
Fig. 203
Example 2. Find the coordinates of the center, the foci, and the ver
tices, and the equations of the asymptotes and the directrices of the
hyperbola
16 x2  9 1/2 4 32 X + 54 ?/  209 = 0.
The given equation may be written in
the form
16(x + l)29(y3)2=144,
or
(a; +1)2 (i/3)2 _.^
9 10
The center is therefore at the point
(—1, 3) and the transverse axis is parallel
to the Xaxis (Fig. 204). Since a = 3, the
vertices are (2, 3) and (4, 3). More
over, since c = V9 + 16 = 5, the foci are
at the points (4, 3) and ( 6, 3). Like
14 4
wise, e = c/a = 5/3 and hence the directrices are x = — — , x =  • The
asymptotes are given by
16(a;+ 1)2 9(2/ 3)2 = 0.
Why? That is, the asymptotes are the lines
4xSy\lS = 0,
and
4xf3y5 = 0.
Y
\ ite "^
^v 4^
_ Cv := ^^
^^^' ^^L
^t" M
A ' /ji\
/ \ '^ / \
._S7_.X
\ ~/^\ uJii
3j::_ _vt/
^Z ^ X2 X
7/^Z__::^g. "
^V ^ i^^^S
Z^ M ^^ \^
7 ^
Fig. 204
352 MATHEMATICAL ANALYSIS [XIII, § 229
EXERCISES
For each of the following hyperbolas determine the transverse axis, the
conjugate axis, the coordinates of the center, the coordinates of the ver
tices and the foci, and the equations of the directrices and asymptotes.
Sketch the curves.
1. Sx^4y^=12. 7.  9 x2 + 2/'^ = 36.
2. 4 x2  3 2/2 = 12. 8. 2/2  2x2 = 4.
3. 4x23?/2z=_ 12. 9. 4x2 12?/2_8x 242/  56 = 0.
4. 3 x2  4 2/2 =  12. 10. 5 x2  4 ?/2 f 10 X + 25 = 0.
5. _36x2+25?/2 = 144. 11. 9x2  16i/2 + 18x  96y  279 = 0.
6. x2  t/2 = 1. 12. x2  2/2 4. 2 X  2 2/ = 2.
13. Write the equations of the following hyperbolas :
(a) Center at origin, transverse axis = 6 on xaxis, conjugate axis = 4.
(&) Center at origin, transverse axis = 8 on yaxis, conjugate axis = 10.
(c) Center at origin, transverse axis and conjugate axis = 4, axes
coinciding with coordinate axes. Two solutions.
(d) Center at origin, focus at (5, 0) and transverse axis = 8.
(e) Center at origin, transverse axis = 8, focus at (0, 5).
(/) Center at Origin, focus at (5, 0), conjugate axis = 8.
(g) Center at (1, 2), transverse axis = 6 parallel to xaxis, conjugate
axis = 4.
(A) Center at (0, 3), focus at (0, 5), conjugate axis = 2V3.
(i) Foci at (1, 2) and (1, — 8), transverse axis = 6.
14. A hyperbola has its center at the origin and its axes on the
coordinate axes; it passes through the points (0, VS) and (2, 3). Find
its equation.
[Hint. Since one point of the hyperbola lies on the ^/axis, the equation
may be assumed in the form I^^, i.e.
62 a2
and a and b may then be determined.]
15. Show that the equation of any hyperbola whose vertex is at the
origin and whose transverse axis is on the xaxis is of the form a'^y =
hH2 ax } .x2). (See Ex. 16, p. 344.)
16. A hyperbola whose asymptotes are at right angles is called rectan
gular. Prove that the equation of a rectangular hyperbola may be written
in the form x2 — w2 = cfi.
XIII, § 231]
THE HYPERBOLA
353
230. Focal Radii of the Hyperbola. If P{xi, y^) is any
point on the hyperbola whose equation is
i^2 _ ^ ^
the focal radii F^P and F2P are given by the equations
F^P = exi + a, F^P = ex^ — a.
The proof of the above statement is left as an exercise. It
is analogous to the corresponding proof ^n the case of the
ellipse (§ 225).
Hence, the difference of the focal radii of any point on a hyper
bola is a constant.
A hyperbola could, therefore, be defined as the locus of
a point which moves so that the difference of its distances from
two fixed points (the foci) remains constant.
231. Conjugate Hyperbolas. Any hyperbola determines
uniquely a second hyperbola whose transverse and conjugate
axes coincide in position and length with the conjugate and trans
verse axes respectively of the first
hyperbola (Fig. 205) . Thus, if the
equation of the first hyperbola is
a2
62
1,
the equation of the second hyper
bola is ^2 2/2
x^
a2
Fig. 205
Each of the two hyperbolas thus related is called the conjugate
of the other, and the two hyperbolas are called conjugate
hyperbolas.
Two conjugate hyperbolas have the same asymptotes. Why ?
2a
354 MATHEMATICAL ANALYSIS [XIII, § 231
EXERCISES
1. Geometric construction of the hyperbola. Show how to construct
a hyperbola given the foci and the length of the transverse axis by a
method depending on the property of the hyperbola derived in §230
and entirely analogous to the first method described in § 226 for con
structing the ellipse.
2. Derive the equation of the hyperbola from the definition suggested
at the end of § 230. [Let the foci be i^i(c, 0) and i^2( c, 0) and let
the constant difference of F\P and F2P be 2 a.]
3. What is the equation of the hyperbola x"^ — y^ — a^ after it has
been rotated about the origin through an angle of 45° ? (Cf. § 190.)
4. From the result of Ex. 3 determine the length of the transverse axis
of the hyperbola xy = k.
5. What are the equations of the hyperbolas conjugate to the hyper
bolas in Exs. l12,p. 352?
6. Prove that the foci of two conjugate hyperbolas are on a circle.
THE PARABOLA
232. Standard Equations of the Parabola. If in § 217 we
let e = 1, equation (2) becomes
(17) y^2px+p^ = 0.
or
(18) y' = 'ip(^fi
We saw in § 183 that this equation repre
sents a parabola whose vertex is at the
point (p/2, 0) and whose axis coincides
with the line y = (Fig. 206). If the curve is moved parallel
to the icaxis so that its vertex coincides with the origin, the
equation of the curve becomes
The focus of the curve is now at the point (i>/2, 0) and its
directrix is the line x = — p/2 (Fig. 207).
Xlir, § 233] THE PARABOLA 355
The following theorems follow directly. Their proofs are
left as exercises.
The equation of a parabola whose vertex is at the origin
and whose axis coincides with the ?/axis is
(I,) x^=2/.j,.
The equation of a parabola whose vertex
is at the point {li, k) and whose axis is
parallel to the a;axis is
(II,) (yky = 2p(xh).
Fj(±l\o) X
Fia. 207
The equation of the parabola whose vertex is at the point
(h, k) and whose axis is parallel to the 2/axis, is
(II,) {xhy = 2piyk).
The equation of any parabola whose axis is parallel to the
ccaxis is of the form
(III.) By'^ + Dx\Ey^C = 0.
The equation of any parabola whose axis is parallel to the
2/axis is of the form
(in J Ax^JrDx\Ey\.C=0,
The distance from the vertex to the focus and from the
directrix to the vertex of the parabola y^ ==2px is p/2.
233. Geometric Properties of the Parabola. The chord
drawn through the focus and perpendicular to the axis is
called the latus rectum. Its length is twice the distance from
the focus to the directrix.
The focal radius connecting any point P{xi, 2/1) on the parabola
y2= 2px to the focus is equal to Xi {p/2.
The proofs of these properties are left as exercises.
356
MATHEMATICAL ANALYSIS [XIII, § 234
234. Illustrative Examples. Example l. Given the parabola
r^ = 6 y. Find the coordinates of tlie vertex and
the focus, and the equation of the directrix.
Sketch the curve.
The vertex is at (0, 0) and the axis of the
curve coincides with the yaxis (Fig. 208). The
distance from vertex to focus is 3/2. Therefore
the focus is at (0, 3/2). Likewise, the distance
from vertex to directrix is 3/2. Hence the equa
tion of the directrix is y =— 3/2. To sketch the
curve, mark the focus, draw the latus rectum and
then sketch the curve.
1
_:_ _ + ^_
\ :
V ~t
L A
^ """*:
y ^
Fig. 208
Example 2. Given the parabola ?/2=:— 8x+2?/+15. Find the coordi
nates of the vertex and the focus, and the
equation of the directrix. Sketch the curve.
The given equation may be written as
(yl)2=8(a:2).
Therefore the vertex is at (2, 1) (Fig. 209),
and the axis is parallel to the a!;axis. The
distance from vertex to focus and from
directrix to vertex is — 2. Therefore tlie
focus is at (0, 1) and the equation of the
directrix is x — 4t. The curve is readily
sketched by plotting the focus and marking
off the latus rectum. It may also be sketched by plotting another point
or two.
1
'H
~
'*'
s
I
s
s
\
(0,
1
J
\:
/
/
/
/
^
T
/
X
'
Fig. 209
EXERCISES
Sketch each of the following parabolas. Determine the coordinates
of the vertex and the focus, and the equation of the directrix.
1. y2 = 4a.,
2. y^ =—4x.
3. 2/2 = 4 ic + 2.
4. y2__4a._^2.
5. x2 = 4 y.
6. x2=4y.
10. a;2 } 4 X — 4 y + 6 = 0.
11. 2/2 _ 2a; 4?/ 8 = 0.
12. a:2 4 r/ ( 1 = 0.
7. x^ = 4y +2.
8. x2 =4?/ + 2.
9. 2/2 = 6x+ 12.
13. 2/2 =  4 X + 2 y f 8.
14. y'^ + 2x4y = 0.
16. x2  2 X } 2 y = 0.
XIII, § 235] THE PARABOLA 357
16. Write the equation of each of the following parabolas :
(a) Vertex at (0, 0) and focus at (2, 0).
(6) Vertex at (0, 0), axis coinciding with ?/axis, curve passing through
the point (8, 4).
(c) Focus at (— 1, 3) and directrix the line a; — 1 = 0.
(d) Vertex at (1, — 2), axis parallel to xaxis, distance from vertex to
focus equal to 2.
(e) Vertex at (0, 2) , directrix parallel to a:axis and parabola passing
through the point (2, 1).
235. The Intersections of Conies and Straight Lines. The
coordinates of the pomts of intersection of the ellipse
(19) 6V f ay = a'^b^
and the straight line
(20) y = ma; + k,
are found by solving these two equations simultaneously for
(x, y). Eliminating y, we obtain the quadratic equation
(21) {W + a}m')x^ + 2 a'mkx + a\k''  h^) = 0,
the roots of which are the abscissas of the points of intersection.
For each of these roots the corresponding ordinate is found by
substituting in (20). Why not in (19) ? We accordingly ob
tain, in general, two solutions {x, y). These solutions are real
and distinct, real and equal, or imaginary, according as
(22) 62'^ (j2^2 _ ]^2 ^0, =0, or < 0.
Corresponding to these three cases, the straight line intersects
the ellipse in two distinct points, in two coincident points {i.e.
in a single point), or not at all.
The discussion just given includes for a = 6 the case of the
intersection of a circle and a straight line.
To treat the intersection of the hyperbola ly^x^ — a V = «^^^
with the straight line (20), we need only notice that alge
braically we can reduce this problem to the preceding by
simply writing — 6^ for h"^. Why ?
358 MATHEMATICAL ANALYSIS [XIII, § 235
This leads to the equation
((^2^2 _ 52)3.2 ^ 2 a^mkx + a2(fc2 + b"^) = 0.
This is a quadratic equation unless a^mS _ 52 _ q^ if ^^2^2 _ 52
= 0, the line (20) is parallel to an asymptote, and, ii k ^ 0, it
meets the hyperbola in only one point. If A: = the line is an
asymptote and does not meet the curve at all. If a'^m'^ — b^z^O,
we conclude that the line (20) intersects the hyperbola in two
distinct points, two coincident points {i.e. in only one point), or
not at all, according as
(23) k'^o?m^\¥>0, =0, < 0.
Finally, the line (20) will meet the parabola
(24) y' = 2px,
in the points whose abscissas are the roots of the equation
m2a;2 + 2(mk  p)x \ k^ = 0.
If m = 0, the line meets the curve in only one point. If m ^ 0,
the line will intersect the parabola in two distinct points, two
coincident points, or not at all, according as
(25) p2mk>0, =0, OY < 0.
Similar results are evidently secured also for straight lines
x = k, parallel to the 2^axis. We then have the theorem :
Any conic is met by a straight line in the plane of the conic in
two distinct points, a single point, or not at all.
EXERCISES
1. Draw figures illustrating all the results of the last article.
2. In a manner similar to that of the last article discuss the intersec
tions of the line y = mx ^ k and the conic y'^ = 2px~ gx^.
3. Derive conditions analogous to (22), (23), and (25) of the last article
when the straight line is assumed in the form Ax + By \ C = 0. These
conditions are slightly more general than those given in the text. Why ?
XIII, § 236] THE PARABOLA 359
236. Tangents and Normals. Slope Forms. If, for a given
value of m, the valae of k in the equation y = mx f k is so deter
mined that the intersections of the line y = mx \ k with a given
conic coincide, i.e. so that the quadratic equation determining
the abscissas of the points of intersection has equal roots, the
line will be tangent to the conic. Why ? (See § 209.)
The slope forms of the equations of the tangents to a conic
result directly from the middle one of each of the conditions
(22), (23), and (25) for the determination of k. Hence the
equation of the tangent whose slope is m is :
for the ellipse b'^x'^ f a^y^ = a^b^^
(26) y =mx± Va^m^ + b^
for the hyperbola b'^x'^ — a^y^ = a^b"^,
(27) y = mx± Va^m^  b^ ;
for the parabola y^ == 2 px,
(28) y^mx+J.
Am
We note that for a given slope the parabola has only one
tangent, the ellipse two, and the hyperbola either two or none
according as ahn^ — b^ '^ or < 0. [The condition a^m^ — &^
= yields the asymptotes.]
The line drawn perpendicular to a tangent through its point
of contact P is called the normal at P.
EXERCISES
1. Find the equations of the tangents to the following conies satisfying
the conditions given, and find for each tangent its point of contact :
(a) 4x2+ 9y2 = 36, m = ^.
(6) y^ = Sx, inclination 30°, 45°, 135°.
(c) 9 x2  25 2/2 = 225, perpendicular to x + y {I =0.
{d) x^ y^ = 1, parallel to 5 x f 3 y — 10 = 0.
(e) y^ = 8x, perpendicular to 2x — Sy + Q = 0.
360 MATHEMATICAL ANALYSIS [XIII, § 236
2. Show that the line y = mx ± Vb'^ — a'^m^ is tangent to the hyper
bola &2a;2 — a2y'2 f a2^2 _ Q for all real values of m lor which h^ — d^nf > 0.
3. For what value of k will the line y =2x + k be tangent to the
hyperbola x2 4 2/24 = 0?
4. Find the coordinates of the points of intersection of the line
3x — w + l=0 and the ellipse x^ + 4 y2 = 55.
6. Find the points of contact of the tangents y = mx± Va^m^ + b'^ to
the ellipse b'^x'^ + a^y^ = a^b'^.
6. From the result of Ex. 5 find the equations of the normals to the
ellipse &2x + a'^y'^ = a'^b'^ wliose slope is m.
7. By the method suggested in Exs. 5 and 6, find the equation of the
normal to the hyperbola bH^ — ay^ = a^b'^ in terms of its slope.
8. Same problem as Ex. 7 for the parabola y^ = 2 px.
9. A tangent to the ellipse b'^x:^ \ a^y^ = ab^ will pass through the
point (xi, yi), if yi = mx\ ± Va'^m'^ + b'^. By solving this equation for m
show that through a given point (xi, yi) will pass two distinct tangents, one
tangent, or no tangents, according as b'^Xi^^a^yi^—a^b'^ > 0, =0, or < 0.
10. By the method of Ex. 9, discuss the number of tangents that can
be drawn from a given point (xi, yi) to the hyperbola b'^x^ — a^y^ = cfib'^',
to the parabola y'^ =2 px.
11. Find the equations of the tangents to the parabola y^ = ix which
pass through the point (—2, — 2).
12. Find the equations of the tangents to the ellipse 4 x^ + y2 = ig
which pass through the points ( V3, 2) ; (0, 4) ; (0, 8) .
237. Tangents. Point Form. The slope of the curve
(29) Ax^ + By^ + Dx + Ey\C= 0,
at a point (a^i, 2/1) on the curve, was found in § 184 to be
2By,\E
Hence the equation of the tangent to (29) at {xi, yi) is
2/ — 2/1 = — — — ^ — (» — ^i)'
This reduces to
(30) 2 Axix + 2 Byiy + Dx\Ey = 2 Ax^'^ + 2 By^^ f Dx^ f Eyi
XIII, § 237] THE PARABOLA 361
Since (xi, yi) is by hypothesis on the curve (29), we have
2 Ax,^ 4 2 Byi^ = 2 Dx^2 Eyi2 C.
Substituting this value in the righthand member of (30),
2 Axx^ + 2 Byiy \ Dx \ Ey =  Dx^  Ey^ 2 0.
Hence, by transposing and dividing by 2, we obtain the equation
of the tangent to (29) at the point (x^ y^ in the form
(31) Ax,x + By,y + 2)(^^±^ + E^^^^ + C = 0.
A A
This equation is readily written down from (29) by replacing
a;2, 2/2, X, and y by x^x., y^y, ^{x + x^), and }(y + y^), respectively.
By applying this rule to the standard equations of the conies
which are special cases of (29) we obtain :
The equation of the tangent at the point (xi, y^
to the ellipse « y2
is
Jfi^, l/i2/_1.
a2 + b2^'
to the hyperbola
aj2 y^ ^
a2 62
is
to the parabola
2/2 = 2px
is
ViV = P{x + Xi
EXERCISES
1. Write the equation of the tangent to each of the following conies
at the point indicated :
(a) x2 + 4i/2 = 8, at (2, 1).
(6) 4a:2_3?/2 = 9, at (3, 3).
(c) y'^ — Qx = 0, at the point where y = — 3.
(d) x'^y'^z=zA^ at (2, 0).
(c) x22?/2^_4, at(2, 2).
(/) y2 — 4 X = 0, at the extremities of the latus rectum.
362
MATHEMATICAL ANALYSIS [XIII, § 237
2. Write the equation of the normal to each of the conies in Ex. 1 at
the point indicated.
3. Find the equation of the normal to each of the conies hH"^ + ahj'^ =
a^b^ bx^  aV = «^&^» and y^ = 2px at the point (xi, yi).
4. Prove that the tangents drawn to an ellipse at the extremities of any
diameter (chord through the center) are parallel.
6. Show that an ellipse and a hyperbola with common foci intersect at
right angles.
6. Show that the tangents at the vertices of a hyperbola meet the
asymptotes in points at the same distance from the center as are the foci.
7. Find the angle (in degrees and minutes) at which the two curves
a;2 + 2 y'^ = 9 and ?/2 + 4 x = intersect.
8. Show that the secant of the parabola y^ = 2px joining the points
(xi, yi) and (X2, yt) on the curve has the equation 2 px  (?/i + yi)y \ yiy2=0.
Show that this reduces to the equation of the tangent when the given
points coincide.
238. Geometric Properties of Tangents and Normals to
the Parabola. Let the parabola have the focus F, the vertex
V, and the directrix d, the latter
meeting the axis VF in D (Fig.
210). If the vertex is chosen
as origin of a system of rectan
gular coordinates and the axis
is chosen as the avaxis, while
the segment DF is denoted by
p, the equation of the parabola
is y^ = 2 px. Now let P{x^, 2/0
be any point on the parabola. The equation of the tangent at
this point is y^y =p{x \x^. This tangent meets the axis of
the parabola (the a>axis) in the point T{—x^y 0). Hence
TV= VM,
where M is the foot of the perpendicular dropped from P on
the axis. From this, and by the definition of the parabola,
d
L
V^^^
/^
'V
\
T D
Fig. 21
XIII, § 238] THE PARABOLA 363
follow the relations
TF=DM=LP=FP,
where L is the foot of the perpendicular drawn from P to the
directrix. Hence TFPL is a rhombus. We conclude further
that ZLPT=ZTPF',
and, if S is the intersection of the diagonals of the rhombus
TFPL, that the angle FSP is a right angle. Moreover, the
line drawn through V, the midpoint of TM, perpendicular to
TM, passes through S. We have then the following theorems :
Theorem 1. Tlie tangent to a parabola at any point P bisects
07ie of the angles formed by the focal radius of P ayid the line
through P parallel to the axis of the parabola ; the normal at P
accordingly bisects the other angle.
Theorem 2. The foot of the perpendicular djopped from the
focus on any tangent to the parabola is on the tangent at the
vertex.
EXERCISES
1. Prove theorems 1 and 2 of § 238 analytically.
2. Give a geometric construction for the tangent to a given parabola
at a given point. (The axis of the curve as well as the curve is supposed
to be given. )
[A geometric construction means a construction with ruler and
compass.]
3. Given the focus and directrix of a parabola, show how any num
ber of points of the parabola can be constructed on the basis of the
results of the last article.
4. Given the focus of a parabola and the tangent at the vertex, use
Theorem 2 of § 238 to draw any number of tangents to the parabola.
These tangents will give a vivid picture of the shape of the curve ; the
tangents are said to envelop the curve. The curve itself is not supposed
to be given.
5. The outline and axis of a parabola are given ; show how to
construct the focus and directrix.
364 MATHEMATICAL ANALYSIS [XIII, § 238
6. To construct the tangents to a giv6n parabola from a given
external point. Assume that the focus and directrix and hence the axis
are given.
[Analysis: If Q is the given point, it
follows from Theorem 1 of the last article
that A QLiPi and A QFPi are congruent.
Hence,
Li
^'
^
^,
kD
KX
l;
'XpF
IV
QLi = QF.
We determine Zi (and Z2), therefore, as
the intersection with the directrix of the
circle with center Q and radius QF. Com
plete the construction. How is the con
struction affected when Q assumes various positions in the plane ? When
is the construction impossible and why ? What happens when Q is on
the curve ?
7. In the figure of Ex. 6, prove that the line through Q parallel to the
axis bisects the "chord of contact" P1P2.
8. If a parabola is rotated about its axis the sur
face generated is called a paraboloid of revolution.
Prove that if a source of light is placed at the focus
of such a paraboloid*, all the rays issuing from the
source will be reflected in the same direction (par
allel to the axis of the paraboloid). This is the prin
ciple of the socalled parabolic reflectors, used on searchlights, etc.
9. By an argument similar to that employed in § 212, prove that the
equation of the chord of contact of the tangents drawn from an external
point (xi, 2/1) to the parabola y' = 2px is ijiy =p(x + Xi). This line
is called the polar of the given point with respect to the parabola. It is
defined by its equation even when no tangents can be drawn through the
given point.
10. Prove that the polar of a point Q is parallel to the tangent at
the point in which the line through Q parallel to the axis meets the
parabola.
11. Prove that the length of the socalled subnormal My oi a parabola
at the point P (see Fig. 210) is independent of the position of P on the
curve.
12. Prove (Fig. 210) that TF = FN = FP and that FS=\ FN.
♦The focus of the generatmg parabolai is called the focus of the paraboloid.
XIII, § 238]
THE PARABOLA
365
13. Use the relation FN' = FP (Ex. 12) to show how to construct the
normal at a given point P of a parabola (the focus and axis also being
given). Construct a considerable number of normals in this way and
show that they envelop a curve. (See Ex. 4 for the meaning of
" envelop.")
14. Show that any two perpendicular tangents to a parabola intersect
on the directrix.
239. Geometric Properties of Tangents and Normals to
the Ellipse. If for any ellipse we let the coordinate axes coin
cide with the axes of the curve, the equation of the ellipse has
the form
The equation of the tan
gent at any point Pi(a;i, i/i) is
b^x^x h a%y = a^b^. _
A
The a>intercept (Fig. 211)
of this tangent is
Xi
The remarkable thing about
this result is the fact that it is independent of b and of y^.
This means that if any other ellipse be given having the axis
A' A in common with the first ellipse, then the tangent drawn
to this new ellipse at a point having the abscissa x^ will also
pass through T. This is therefore true of the circle drawn on
A' A as diameter. If A' A is the major axis of the ellipse, this
circle is called the major circle of the ellipse ; similarly the
circle drawn on the minor axis of any ellipse as diameter is
called the minor circle.
* We do not in this article impose the restriction a > 6.
366 MATHEMATICAL ANALYSIS [XIII, § 239
A geometric construction for the tangent at any point Pj of
an ellipse follows readily from the above considerations (as
suming that in addition to the curve one of the axes is given).
Figure 211 shows the construction using the major circle and,
in broken lines, the construction using the minor circle.
The following theorem is of fundamental importance in dis
cussing the geometric properties of the ellipse :
Theorem 1. Tlie tangent and the normal to an ellipse at a given
point bisect the angles formed by the focal radii drawn to the point.
Proof. We are to prove that the tangent at Pj (Fig. 212)
bisects the angle F^PxR, and that the normal at P^ bisects the
angle F^P^F^. To this end we calculate first the tangent of
the angle SP^R. Using the equation of the ellipse as given
above and taking the foci to be PgCc, 0) and Pi(— c, 0), we have
the slope of the tangent P^S =  ^,
a^y^
the slope of F,R (i.e. FiP,) = ^ —
The tangent of the angle <^i from P^S to P^R is then
Xi \ c a^yi
tan d)i — —
^' 1 b%y,
a%{x, + c)
Simplifying this expression, we find
tan <^i = — .
The tangent of the angle <^2 from P^S to P^F^ may evidently
be obtained by simply changing c to — c in the last result.
(Why?) Hence, .,
tan <f)2== •
XIII, § 239]
THE PARABOLA
367
We conclude that <^2 = — ^v This proves that PiS bisects the
angle F^PiB. That the normal bisects the angle FiP^F^
follows at once from elementary geometry.
The theorem just proved leads at once to another geometric
construction for the tangent (and normal) to an ellipse at
a given point, supposing the foci of the ellipse are known.
Theorem 2. The foot of the perpendicular dropped from
either focus on any tangent to an ellipse lies on the major circle.
Proof. (See Fig. 212.) Let S be the foot of the per
pendicular dropped from F^ on the tangent P^S, and let it
meet the line F,P^ in R. Then F^^P^B
is an isosceles triangle (why?) with
P^R = P.F^. We have then
F,R = F,P, + P,F^ = 2 a. (§ 225)
Also aS' is the midpoint of F2R and
O is the midpoint of FiF^. Hence
OS=^FiR==a, and S is on the major
circle.
We should note also that, if Q is any
point on the tangent PiS, then QR = QF2, which is important
in connection with the problem of drawing the tangents to an
ellipse from an external point. (See Ex. 5, below.)
Fig. 212
EXERCISES
1. Show how to construct the tangent to a given ellipse at a given
point. (Two constructions, one using the major circle, one using the
foci.)
2. Show that, in Fig. 211, OA is a mean proportional between OM
and OT.
3. Show that, in Fig. 212, OFi is the mean proportional between the
intercepts on the ccaxis of the tangent and normal at Pi.
368 MATHEMATICAL ANALYSIS [XIII, § 239
4. Prove analytically that S (Fig. 212) is on the major circle.
5. Show how to construct the tangents to an ellipse from a given ex
ternal point Q. [Hint : Construct B (Fig. 212) as the intersection of two
circles, one with center Fi the other with center Q.]
6. Show that if a right angle moves with its vertex on a given circle
and one of its sides passing through a fixed point within the circle, the
v^ther side will envelop an ellipse.
7. Use the result of Ex. 6 to construct a considerable number of tan
gents to an ellipse, given the major circle and one focus (the outline of
the ellipse is not supposed to be given in advance, but will appear vividly
after this problem is solved).
8. If an ellipse is rotated about its major axis the surface generated is
called a prolate spheroid. Show that sound waves issuing from one focus
will be reflected by the surface to the other focus. This principle is used
in the socalled ' ' whispering galleries. ' '
9. By an argument similar to that used in § 212 show that the equation
xxi/a^ f yyi/b'^ = 1 is the equation of the line joining the points of contact
of tangents drawn from (xi, yi) to the ellipse x^/a''^ + y'^jW — 1.
240. Geometric Properties of the Hyperbola. Many of
the geometric properties of the hyperbola are similar to cor
responding properties of the ellipse, which is to be expected
in view of the similarity of their equations. The following
two theorems are fundamental :
Theorem 1. Tlie tangent at any point of a hyperbola bisects
the angle betiveen the focal radii diaion to the point. The normal
bisects the adjacent supplementary angle.
Theorem 2. The foot of the perpendicular dropped froin
either focus on any tangent to a hyperbola is on the circle drawn
on the transverse axis as diameter.
The proofs of these theorems are left as exercises. They
are similar to the proofs of the corresponding theorems on the
ellipse. Draw figures illustrating Theorems 1 and 2.
XIII, § 240] THE PARABOLA 369
Certain new properties of the hyperbola relating to the
asymptotes will be found among the exercises below..
EXERCISES
1. Show how to construct the tangent and the normal to a given
hyperbola at a given point.
2. If P is any point on a hyperbola, OA the semi transverse axis,
ifcf the foot of the perpendicular dropped from P on OA (produced), and
T the point in which the tangent at P meets OA, prove that OA is a
mean proportional between Oilf and OT.
3. With the notation of Ex. 2 show that OFi is the mean propor
tional between ON and OT, Fx being the focus on OA and N the point
in which the normal at P meets OA (produced).
4. Prove Theorem 2 (§ 240) analytically.
5. Show how to construct the tangents to, a hyperbola from an ex
ternal point.
6. Show that if a right angle moves with its vertex on a given circle
and one of its sides passing through a fixed point outside the circle the
other side will envelop a hyperbola.
7. The construction of tangents to a hyperbola analogous to Ex. 7,
p. 368.
8. Use Ex. 3 above and Ex. 3, p. 367, to show that an ellipse and
hyperbola having the same foci intersect at right angles ,
9. Prove that, if a tangent to a hyperbola meets the asymptotes in
Ti and T2, the point of contact of the tangent is the midpoint of the
segment T1T2.
10. Prove that the area of the triangle formed by any tangent and the
asymptotes of a hyperbola is constant (= a6).
11. Show that if a straight line cuts a hyperbola in Pi and P2 and the
asymptotes in ^1 and ^2 the segments PxQx and P2^2 are equal. Use
this result to construct any number of points of a hyperbola when the
asymptotes and one point of the curve are given.
12. By an argument similar to that used in § 212 show that the
equation xxx/a^ — yyi/h^ = 1 is the equation of the line joining the
points of contact of the tangents drawn from (cci, 2^1) to the hyperbola
a;2/a2  2/2/62 = 1.
2b
370 MATHEMATICAL ANALYSIS [XIII, § 241
241. The Conies as Plane Sections of a Cone. We stated
in § 216 that the ellipse, hyperbola, and parabola could all be
obtained as the plane sections of a right circular cone. This
we shall now proceed to prove. In doing so we shall get the
machinery for solving problems of a more general type.
If a point P in a plane a (Fig. 213) is joined to a point
S not in a by a straight line SP, the intersection P' of SP
by a plane a' is called the projection
of P from S upon a'. Similarly, if
all the points of a curve in a be
joined to S, the intersections of these
lines with a plane a' form a curve C",
Fig. 213 ^^v/ which is called the projection from S
of the curve C. The point JS is called the center of projection,
and the process described is called central projection, to dis
tinguish it from orthogonal projection previously considered
{e.g. in § 135).
If, now, the curve C in the plane a is a circle, the lines
through jS and the points of this circle form a cone with vertex
S. This is not a right cone, in general. As the lines through
S are not supposed to terminate in S, we get a socalled co7n
plete cone, or cone of two nappes, which consists of two con
gruent ordinary cones placed vertex to vertex so that their
axes form a straight line. It will now be clear that a plane
section of this cone is the same as the projection of the circle
C from the vertex S upon the plane of section.
We have then reduced the problem to that of finding the
central projection of a circle. We will solve it by finding the
relation between the coordinates of a point P in a and the co
ordinates of the corresponding point P' in a'. To this end
(Fig. 214) let be the foot of the perpendicular dropped from
JS on the line of intersection of the planes a and a'. Let be
XIII, § 241]
THE PARABOLA
371
the origin and let the line of intersection OF of the two planes
be the 2/axis in the system of coordinates in each of the two
planes. Let the line OX perpendicular to 01^ in a be the
a^axis in a, and the line OX' perpendicular to OF in a' be the
icaxis in a'. Let the angle between the two planes be ; then
X'OX= 0. Now let P{x, y) be any point in the plane a, and
let P\x\ y') be the projection of P from S. We seek the
relation connecting the coordinates x, y, x', y'.
Draw ST parallel to OX, and represent the length OS by h.
Then we have
T0=^, TS= ^
sinO
tand
We then have from similar triangles
x' :{TOix') =x:TS,
y':y = SM^ : SM= TM' : TO.
If we substitute the values of TO, TS, and TM' (= TO + x'),
we obtain ,
x' +
h
h '
sin tan
372 MATHEMATICAL ANALYSIS [XIII, § 241
and , , h
sind
1^
y _Jl_
sin B
Solving these equations for x and ?/, respectively, we have
_ li cos 6 • x' _ hy'
^~ ^mO'X' + h' ^"sin^x' + Zi*
If these expressions be substituted for x and y in the equa
tion of any curve in the plane a, the resulting equation in x'
and y' will be the equation of the projection of the curve in a'.
To solve the problem we proposed at the outset, let the curve
in the plane a be the circle
x'^\y^=a\
The equation of the corresponding curve in a' is then
7^2 cos2 ^ . x'2 f hhj"^ = tt2 sin2 ex'^+2 lia^ sin • x' \ a%\
Collecting like terms, we have *
(/i2 cos2 ea" sin2 6) x" + li'y'^  2 ha" sin 6  x'  aVi^ = 0.
We see at once that this is the equation of a conic. It is an
ellipse, a parabola, or a hyperbola according as
/i2cos2^a2sin2^>0, = 0, or < 0,
i.e. according as
tan ^   < 0, = 0, or > 0.
a
But h/a is the tangent of the angle <^ which an element of the
cone with vertex S makes with the plane a. If is less than
this angle (f>, the section of the cone is an ellipse ; if ^ is equal
to <^, the section is a parabola ; and if 6 is greater than <^,
the section is a hyperbola. Note that this result is in accord
ance with our geometric intuition of the situation.
XIII, § 242] THE PARABOLA 373
EXERCISES
1. Prove that the central projection of any circle is a conic ; that is,
that a plane section (not through the vertex) of any circular cone (not
necessarily a right cone) is a conic.
[Hint : Complete generality will be secured by taking the equation of
the circle in a to be x'^ ^ y'^ {■ dx \ c = 0. Why ?]
2. Prove that the central projection of any conic is a conic.
,3. Prove that the central projection of a straight line is a straight line.
4. Prove that there exists in a just one straight line which has no
corresponding line in a', namely the line of intersection of a with the
plane through S parallel to a'. This line is called the vanishing line of a.
5. Prove that the central projection of a circle in a is an ellipse, a
parabola, or a hyperbola, according as the vanishing line in a meets the
circle in no points, one point, or two points.
242. Poles and Polars. Diameters. We have had occasion
in several exercises to note that the equation which represents
the tangent to a conic at the point Piix^ , y^ when Pj is on the
curve, represents a straight line called the polar of Pi when Pj
is any point in the plane. Pi is then called the pole of the
line with respect to the conic. The polar of a point on the
conic is then the tangent at the point. We have also seen
that the polar of a point Pi through which pass two tangents
to the conic is the line joining the two points of contact of the
tangents. In the more extensive geometric theory of conies
poles and polars play an important role.
A straight line passing through the center of an ellipse or
hyperbola is called a diameter of the conic. Every diameter
of an ellipse meets the curve in two points ; some of the
diameters of a hyperbola meet the curve in two points. These
points are then called the extremities of the diameter, and the
distance between them is called the length of the diameter.
Any line parallel to the axis of a parabola is called a diameter
of the parabola. Other properties are given in exercises below.
374 MATHEMATICAL ANALYSIS [XIII, § 242
MISCELLANEOUS EXERCISES
Properties of Poles and Polars
1. Write the equation of the polar of each of the following points
with respect to the conic given, and draw the corresponding figure :
(a) (1,5); 2a;2 + y2^4. (d) ( 1, 3) ; a;2 + 2/2 + 4 x  6 y 2 = 0.
(6) (2,0); 4x292/2 = 36. (c) (2,  3) ; 2/2 = 6x.
2. Find the pole of the line 3x — 4y412 = with respect to tho
following conies :
(a) 3x2 + 4 y2 = 12 ; (ft) a;2  5 y2 = 20 ; (c) y^ = Sx; (d) x'^ = iy.
3. Prove that in any conic the polar of a focus is the corresponding
directrix.
4. Prove that in any conic, if Pi and P2 are two points such that the
polar of Pi passes through P2, the polar of P2 will pass through Pi.
5. From the result of the last exercise follows geometrically the follow
ing theorem : If a straight line be revolved about a point P and tangents
are drawn at the points where it meets a conic, the locus of the intersec
tion of these pairs of tangents is the polar of P with respect to the conic,
or a part of the polar. Which part will it be ?
6. Prove that the polar of any point on a directrix of a conic passes
through the coiTesponding focus. [See Exs. 3 and 4.]
7. A straight line through a point Pi meets a conic in Ci and C2, and
the polar of Pi in Q. Prove that Pi and Q divide the segment C1C2 in
ternally and externally in the same ratio.
[Solution : Let Pi(xi, yi) and P2(X2, 2/2) he any two points. Then
the point P whose simple ratio with respect to Pi and P2 is X has the co
ordinates ^ _ xi + Xx2 , _ 2/1 + X1/2
If these be substituted in the equation of the ellipse bH^ 4 a^y^ = ajb^ and
the resulting equation arranged as a quadratic in X, we have
/^4.y22_l^^2 + 2f^2^M_2_l^x + f^' + ^'lUo.
V a2 ^ 62 ) ^ \ a^ ^ b^ J \a^ b^ J
The roots of this equation are the simple ratios of the points Pi and P2,
respectively, with respect to the points d, C2 in which the line P1P2 meets
the ellipse. If the segment Ci, C2 is to be divided internally and ex
ternally in the same ratio the roots Xi, X2 of this equation must be equal
numerically, but opposite in sign, i.e. Xi f Xo must be zero. The coefficient
XIII, § 242] THE PARABOLA 375
of X in the above equation would then be zero, if the theorem to be proved
is true. But the condition
«2 ^ 62
is precisely the condition that P'z{x2^ y2) be on the polar,
of Pi with respect to the ellipse. The similar proofs for the hyperbola and
parabola are left as exercises.]
8. Two points P, Q on the line joining two given points d, C2, are
said to divide the segment Gi Ci, harmonically, if they divide the seg
ment internally and externally in the same ratio (i.e. if CxP/PCi =
— O1Q/QC2). Show that the result of the last exercise leads to the fol
lowing : The locus of a point Q^ such that a given point P and the point
Q divide harmonically the segment joining the points in which the line PQ
meets a conic is the polar of P with respect to the conic, or a part of the
polar. Which part is it ? (Compare with the similar question in Ex. 5.)
Properties of Diameters
9. ProTB that the locus of the midpoints of the chords of a conic
drawn parallel to a given chord is a diameter of the conic.
[Solution for the Ellipse : Let the equation of the ellipse be
b^2c^ + a^y^ = a^b'^, and let the slope of the given chord be m. Then any
chord parallel to the given chord isy = mx + k. The abscissas Xi, X2 of
the points in which this chord meets the ellipse are the roots of the
equation
(62 + a2^2)a;2 + 2 a^mkx + a^(k^  62) = 0.
The sum of the roots of this equation is
Xi}X2= —
62 I a^n^^
The coordinates (x', y') of the midpoint of the chord are then
X' = itixi f X2) =  TT^^, y' = mx' +k= ^'^
62a2m2' 62 + a2m2
The coordinates x', y' then satisfy the equation y = — (b^x^/(a^m), no
matter what the value of k is. The locus of the midpoints of the
chords whose slope is m is, therefore, the straight line whose equation is
y = — (62x)/(a2m). Since this straight line passes through the center of
376 MATHEMATICAL ANALYSIS [XIII, § 242
the ellipse, the theorem is proved for the ellipse. The similar proofs for
the hyperbola and parabola are left as exercises.]
10. From the result of the last exercise, show how to construct a
diameter of any conic, and hence (in case of ellipse and hyperbola) how
to find the center, when only the outline of the curve is given.
11. Having given the outline of an ellipse or hyperbola, construct the
center. Then show how to construct the principal axes (make use of the
fact that the principal axes are axes of symmetry ; a circle drawn with
the center of the conic as center and suitable radius will meet the conic in
the four vertices of a rectangle whose sides are parallel to the principal
axes). Then construct the foci and the directrices.
12. Having given only the outline of a parabola, show how to construct
the axis, the focus and the directrix.
13. Show that the tangent drawn to a conic at an extremity of a
diameter is parallel to the chords which the diameter bisects.
14. If two diameters of a conic are such that each bisects the chords
parallel to the other, the diameters are said to be conjugate; and each
is called the conjugate of the other. Prove that if wi, W2 are the slopes
of two conjugate diameters of the ellipse h'^x'^ + a'^y'^ = d^h'^, then we have
m\m2. = — ly^/a^.
15. Prove that, if wi, mi are the slopes of two conjugate diameters
of the hyperbola fe^x^ _ a'iyi — ah'^, we have WiW2 = h'^/a^.
16. The only conic for which all pairs of conjugate diameters are per
pendicular is the circle.
17. The polars of the points on any diameter of an ellipse or hyperbola
are parallel to the conjugate diameter.
18. If one extremity of a diameter of an ellipse Jfix^ \ a^y'^ = aV)^
has the coordinates (xi, yi), one extremity of the diameter conjugate to
the given one will have the coordinates (— yia/b, X\b/a).
19. The area of a parallelogram circumscribed about an ellipse whose
sides are parallel to two conjugate diameters is constant and equal to 4 db.
20. Prove that, if a\ and &i are the lengths of two conjugate semi
diameters of an ellipse, a\^ + fti^ = a^ f h^.
21. Prove that any pair of conjugate diameters of the hyperbola
6%2 _ 052^2 _ 0,252 are also conjugate diameters of h'^x^ — a^y'^ = — aP'h'^.
22. If a diameter of a hyperbola with center meets the hyperbola in
P and the conjugate diameter meets the conjugate hyperbola in Q, prove
that 0P2 _ 0q^ = (£^  62.
CHAPTER XIV
POLAR COORDINATES
243. Review. Polar coordinates, introduced in §§ 112114,
are often useful in studying geometry analytically. The
present chapter is devoted to illustrating some of the principles
involved and their applications.
EXERCISES
1. What is the locus of points for which p is constant ?
2. What is the locus of points for which 6 is constant ?
3. Show that the points (p, d) and (/s, — 0) are symmetric with respect
to the polar axis.
4. Show that the points (/>, 6) and (— p, 6) are symmetric with respect
to the pole.
6. Show that the points {p,.d) and (/>, d + 180°) are symmetric with
respect to the pole.
6. Find the distance between the points A (2, 45 '^) and B(7, 105°).
[Hint. Use the law of cosines.]
7. Prove that the distance between the points (pi, ^i) and (p2, ^2) is
Vpi'^ + pi^  2 P1P2 cos {62 — di).
244. Locus of an Equation. The locus of an equation in the
variables p and is such that :
(1) Every point whose coordinates (p, 0) satisfy the equation
is on the locus or curve, and
(2) A set of coordinates * of every point on the locus or curve
satisfies the equation.
* Not necessarily every set. Thus, the point (2, 60°) = (  2, 240°) is on the
locus of p = 1 + 2 cos 6 ; but the second set of coordinates does not satisfy the
equation.
377
378
MATHEMATICAL ANALYSIS [XIV, § 244
The curve may be sketched by computing a table of corre
sponding values of p and 6, plotting the corresponding points,
and then sketching the curve through them. The amount of
work may often be shortened if one makes use of the following
obvious rules for symmetry :
If a polar equation is left unchanged,
(a) ichen 6 is replaced by — 6, the locus is symmetric with re
spect to the polar axis.
(b) when p is replaced by — p, the locus is symmetric with re
spect to the pole.
(c) ivheyi 6 is replaced by 180° + 6, the locus is symmetric with
respect to the j)ole.
(d) ivhen 6 is replaced 6?/ 180° — 9, the locus is symmetric ivitJi
respect to the line through the pole perpendicular to the polar axis.
It should be borne in mind, however, that none of these rules
are necessary conditions for sym
metry. Why not ?
245. Illustrative Examples.
We shall illustrate the methods
of plotting curves in polar coordi
nates by the following examples.
Example 1. Discuss and plot the
locus of the equation p = 4 cos 6.
The locus is symmetric with respect
to the polar axis. If we plot points
from 0° to 90°, we obtain the upper
half of Fig. 215. Then by symmetry
we obtain the complete graph given in
Fig. 215. Why ?
Fig. 215
d
0°
30°
45°
60°
90°
p
4
3.5
2.8
2
XIV, § 245]
POLAR COORDINATES
379
Example 2. Discuss
and plot the locus of the
equation p — 4t &m'^ 6.
The locus is symmetric
with respect to the pole,
the polar axis, and the Ime
through the pole perpendic
ular to the polar axis. If
we plot points in the range
^=0° to 0=90°, and make
use of symmetry, we have
the complete figure which
is given in Fig. 216.
e
p
0°
30°
1
45°
2
C0°
3
90°
4
Fig. 216
The branches constructed
by symmetry should be
checked by substituting in
the original equation the
coordinates of at least one
point on each branch. Seri
ous errors may thus be
avoided.
Example 3. Discuss and
plot the locus of the equa
tion p =: 1 h 2 cos (9.
The curve is symmetric
with respect to the polar
axis. If we plot points
from 6 = 0° to 6= 180°, we
get the points shown in
Fig. 217. Then by symmetry we get the complete graph, or the curve
in Fig. 217.
Fig. 217
6
0°
30° 45°
60°
90°
120°
135° 150°
180°
P
3
14V3
1 H\/2
2
1
1V2
1V3
1
380 MATHEMATICAL ANALYSIS [XIV, § 245
EXERCISES
Are the following loci symmetric with respect to the pole ? The polar
axis ? The line through the pole perpendicular to the polar axis ?
1. p = a cos 0.
5.
p = a cos 2 d.
9.
p = a sin2 6.
2. p = a sin d.
6.
p'^ = a sin 2 6.
10.
P = sin^l
p = e.
3. p = a (1 — cos
d).
7.
p = a cos 2 ^.
11.
4. p = a(l — sini
9).
8.
p = a sin 2 ^.
12.
p2 cos 0=4.
Discuss and plot
, thel
ocus
of each of the following <
equations.
13. p = 5.
22.
p cos ^ = 4.
31.
p =z 1 + 2 sin e.
14. p= 5.
23.
p cos ^ = — 4.
32.
p = 1 — 2 cos d.
15. p2 = 25.
24.
p sin ^ = 5.
33.
P = 1 — 2 sin 0.
16. ^ = 30°.
25.
p sin ^ =  6.
34.
p = 2 + cos 0.
17. e= 30°.
26.
p = 1  cos ^.
35.
p = 2 + sin d.
18. p = 8 cos d.
27.
p = 1 + cos ^.
36.
p = 4 tan d.
19. p =  8 cos e.
28.
p = 1 ~ sin ^.
1
29.
p = 1 + sin 0.
37.
p — '■
20. p = 8 sin e.
lcos^*
21. p = — 8 sin e.
30.
p = 1 + 3 cos ^.
246. Standard Equations. We shall now derive polar equa
tions for the straight line and the conic
sections.
The straight line. Let CD be any straight
line (Fig. 218) ON = p the perpendicular
let fall upon it from the pole 0, and a the
angle which this perpendicular makes with
the polar axis. Let (p, 0) be any point on
the line.
ON
Then  = cos (0 — a) or cos (a — 6).
But by § 120,
cos (0 — a) = cos (a — $).
Hence
(1) P = p cos (6  a)
is the desired equation.
XIV, § 246] POLAR COORDINATES ' 381
If the line is perpendicular to the polar axis, its equation is
p cos ^ = p. Why ?
If the line is parallel to the polar axis, its equation is
p sin ^ = p. Why ?
The circle. Let C (c, a) be the center of a circle of radius r
and P (pj 6) any point on the curve (Fig. 219). In the triangle
Fig. 219
OOP, OC=c, OP = p and the angle COP = ± (0  a)
depending upon the position of the point P. But since
cos (0 — a) = cos {a — 0), we have from the law of cosines,
§126,
(2) r2 = c2 + p2 _ 2 c p cos (6  a)
as the equation of the desired locus.
If the center C is upon the polar axis (a — nir), equation (2)
becomes
(3) r2 = c2 f p2 ± 2 c p cos d.
If the circle passes through the pole (r = ± c), equation (2)
becomes
(4) p = ± 2rcos(^a).
If the center C is upon the polar axis (a = 0) and the circle
passes through the pole (c = ± r), equation (2) becomes
(5) p = ± 2 r cos 6.
If the center is at the pole (c=0), equation (2) becomes p= ±r.
382
MATHEMATICAL ANALYSIS [XIV, § 246
The Polar Equation of any Conic. The polar equation of
any conic may now be derived. Let DD' be the directrix, F
the corresponding focus, and e
the eccentricity. Let the per
pendicular through F to DD'
meet DD' in L. Let the segment
LF = p, and take F as the pole
and the extension of the line LF
as the polar axis. If P(p, 6) is
any point on the curve and PS
is the perpendicular from P to
DD\ then by the definition of a
PS,
Fig. 220
conic, we have
that is,
or
(6)
FP==e
p = e{p + p cos 6),
1 — e cos 6
which is the polar equation of a conic. If e < 1, the equation
represents an ellipse ; if e = 1, a parabola ; e > 1, a hyperbola.
EXERCISES
1. Derive the equation p = 2 r cos [ (5) , § 246] directly from a figure.
2. Derive tiie polar equation of the ellipse assuming the righthand
focus as the pole and the major axis as the polar axis.
3. Derive the polar equation of a hyperbola assuming the righthand
focus as the pole and the transverse axis as the polar axis.
4. Derive the polar equation of the circle which passes through (0, 0°)
and has its center at (r, 90°) ; (r, 270^).
6. Derive the polar equation of the parabola assuming the focus at the
pole and the directrix the line p sind = p ; psind = — p.
6. The difference of the focal radii of a certain hyperbola is 3, and the
distance between the foci is 6. Find a polar equation of the curve.
XIV, § 247]
POLAR COORDINATES
383
247. Other Curves. What is the advantage of polar coordi
nates ? Why not continue to use only rectangular coordinates ?
The answer to these questions is that in certain kinds of
problems polar coordinates are much more convenient. The
following examples will illustrate the desirability of polar
coordinates.
The limacon. Through a fixed point upon any given circle
of radius a, a chord OPi is drawn and produced to P so that
Fig. 221
P^P = kj where A; is a given constant (Fig. 221). Find the
locus of P"* as Pi describes the circle.
If p — 2 a GosO is the
equation of the circle and
the pole is the fixed point,
then the locus of P is
(7) p = 2 a cos 0\Jc.
If k = 2 a, the equation
may be written in the form
(8) p = 2 a(l + cos 0)
and the curve is known as the
cardioid, on account of its
heartshaped form (Fig. 222). fig. 222.  The Lima9on
* This curve is known as the lima<^on of Pascal. It was invented by
Blaise Pascal (16231662) , a famous French mathematician and philosopher.
The word lima(;on means snail. The Germans call this curve die PascaVsche
Schnecke.
384
MATHEMATICAL ANALYSIS [XIV, § 247
The cissoid. OA is a fixed diameter of a fixed circle (Fig.
223). At the point A a tangent is drawn, while about the
point a secant revolves which meets the tangent in B, and
the circle in C. Find the locus of a point P on OB so deter
mined that 0P= CB.
Fig. 223
Take as the pole and OA as the polar axis of a system of
polar coordinates. If we denote OA by 2 a, then
the equation of the circle is p= 2 a cos 6. Let P
be denoted by (p, 6). Now
p=OP=OBPB.
0B=2aseGe and PB = 0C=2acose.
p = 2 a (sec — cos 0),
p = 2 a tan ^ sin 0.
Fig. 224. — The
Cissoid The locus of this equation is given in Fig. 224.
The curve is known as the cissoid of Diodes.*
* Cissoid (Greek, Ktoods = ivy) means ivylike. The Greeks considered
only the part of the curve lying within the circle. Diocles was a Greek
mathematician who lived sometime between 217 b.c. and 70 b.c. By means
of this curve, Diocles showed how to construct the side of a cube whose
volume is twice the volume of a given cube. See Ex. 4, p. 388.
XIV, § 247]
POLAR COORDINATES
385
Conchoid of Nicomedes.* A straight line revolves about a
fixed point and meets a fixed straight line MN in the point
Q. From Q a fixed length is laid off on OQ in both direc
tions. The locus of the two points, P and P', thus determined
is called a conchoid.
Let be the pole and the line OR through perpendicular
to MJSf be the polar axis of a system of polar coordinates
Fig. 225. —The Conchoid
Let (p, 6) be the coordinates of any position of the generating
point P (or P'). Then
p = OP(or OP') = OQ±QP=OBseGe± QP.
But OR and QP are given constants ; call them a and b
respectively. Then
(10) p = asec^±6
is the desired equation of the conchoid.
* Conchoid (Greek, Koyxo? = mussel) means mussellike. Nicomedes was a
contemporary of Diodes. He invented the conchoid for the purpose of trisect
ing an angle, which is one of the famous problems of antiquity. This prob
lem cannot be solved by means of the compass and straightedge alone.
2c
386
MATHEMATICAL ANALYSIS [XIV, § 248
248. Spiral Curves. A spiral is a curve traced by a point
which revolves about a fixed point called the center, but con
tinually recedes from or continually approaches the center ac
cording to some definite law.
The spiral of Archimedes is the locus of a point such that its
radius vector is proportional to its vectorial angle. Therefore
its equation is
(11) p = ke,
where Ic is a constant.*
The form of the equation shows that the locus passes through
the pole, and that the radius vector increases without limit as
the number of revolutions increases without limit. Figure 226
represents a portion of the locus for k = ^j with $ expressed
in degrees.
The hyberbolic or reciprocal spiral is the locus of a point
such that its radius vector is inversely proportional to its vec
* In this example, and in those that follow, it is usual to express the angle
in radians ; but this is not necessary, since the same result can be obtained by
choosing a different value for A: if d is exi>ressed in degrees.
XIV, § 248]
POLAR COORDINATES
387
torial angle. The equation of the locus is therefore
(12) P=^.
where A; is a constant.
Figure 227 represents a
portion of the graph for
A; == 70 and for positive
values of 6 (expressed in
degrees).
The logarithmic spiral
is the locus of a point ^^^ ^27
such that the logarithm of its radius vector is proportional to
its vectorial angle, i.e.
(13) log p = kO,
where A; is a constant. If the base of the system of logarithms
is b, the equation may be written in the form p = 6*^. Figure
228 represents a portion of this locus when 6 = 3, for A: = y^,
with 6 expressed in degrees.
388
MATHEMATICAL ANALYSIS [XIV, § 248
EXERCISES
1. Discuss the form of the limagon (7), when  A:  < 2 a. When
A:>2a.
2. Solve the locus problem used to define the lima9on by means of rec
tangular coordinates, and compare the merits of the two solutions.
3. By taking the line OA (Fig. 223) as the icaxis and the tangent to
the circle at O as the yaxis, prove that the equation of the cissoid is
2ax
4. Duplication of a Cube. In the adjoining figure, let MN= a and
MB = 2 a. Draw BA and let it meet the cissoid
in the point D whose ordinate is LD. Prove
that ZZ)3 = 2 OLK If MB = n • a prove that
LD^ = n ■ OL^.
6. If in Fig. 225 the line MN^ is taken as the
a;axis and the line OB A as the jzaxis, prove that
the equation of the conchoid is
a:V=(y + «)'(&' 2/2).
Compare the merits of this solution with that on
p. 385.
6. Trisection of an angle. Let AOB be the angle to be trisected.
Through a convenient point A on one side OA of the angle draw AB per
pendicular to OA. Through
B draw a line B C parallel to
OA. From as fixed point,
and AB as fixed line, and
2 . OB as a constant dis
tance, describe a conchoid
meeting BC in C. Angle
^OC is then \ AOB.
[Hint: E is the midpoint
of Z>C ; then OB = BE = EC. The result then follows from elementary
geometry.]
7. Show that in the conchoid, if
(a) 6 > a, the curve has an oval at the left, as in Fig. 226;
(6) 6 = a, the oval closes up to a point ;
(c) 6 < a, there is no oval and both branches lie to the right of the
point 0.
XIV, § 249]
POLAR COORDINATES
389
8. Draw the parabolic spiral, which is defined by the equation p^ = kd.
Take k = ^^ with 6 in degrees, and use only the positive values of p.
9. Draw the lituus, which is defined by the equation p^ = k/d. Take
A; = 90 with 6 in degrees, and use only the positive values of p.
249. Relation between Rectangular and Polar Coor
dinates. Take the origin of a system of rectangular axes
as the pole, and the positive half of the a;axis as the polar
axis, of a system of polar coordinates.
Let (x, y) and (p, 6) be respectively the rectangidar and polar
coordinates of any point P. Then x/p — cos ^, yjp — sin 6.
Hence, we have
I X = p cos e,
I y = p sin e.
(14)
It is here assumed that the coordinates of P are so chosen that
OP — p and angle XOP = 0. This is always possible. If p is
positive, X always has the sign of cos and y the sign of sin $,
ft<=^' ?^
F P
Fig. 229
Conversely, if p is positive, we see from Fig. 229 that
.2 _ ^2 _L „2 gin $ — ^
(16)
x" + y^,
$ = arc tan [t\ cos d = ■' '
390 MATHEMATICAL ANALYSIS [XIV, § 249
EXERCISES
Transform the following equations into equations in rectangular coordi
nates. State in each case whether the graph is easier to sketch from the
polar or the rectangular equation.
1. p = 1  cos 0. 6. p2 sin 2 e = 3. 11. p^ = 6.
2. p=l + sin^. 7. p2cos2^ = 4. ^^ ^2^1
3. p = 4.
8. p2 = cos 2 e.
13. p = a sec d\ b.
4. pcos^ = 5. I 14. p = 2asec^tan^.
10. p = — •
6. p sin ^ =  2. e 15. p = 4 cos 2 6.
Transform the following equations into equations in polar coordinates :
16. ^2 + 1/2 = 4 X. 21. xy = 4.
17. (x2 + y2)2 _aj2 _ 2/2. 22. a; cos a 4 y sin a = ;>.
18. X y = 0. 23. (y2 4 x2 _ 2 a;)2 = x2 f y2.
19. 1/2 = 4 X. 24. x3 = ?/2(2_a;).
20. 9 x2 + 4 ?/2 = 36. 25. x2?/2 = (y + 2)2(4  y^).
MISCELLANEOUS
EXERCISES
Sketch the follow
fving
curves :
1. p = a cos 2 d.
1. p = a cos 6 6.
2. p = a cos 8 d.
8. p = a sin 5 ^.
3. p = a sin 2 ^.
4. p = a sin 3 6.
• 6
9 p = a sin
h. p = a cos 4 ^.
6. p = a sin 4 ^.
10. p = acos
Find the points of intersection of the following pairs of curves. Plot
the curves in each case and mark with their respective coordinates the
points of intersection.
11. p == 1 4 cos e,. 14. p = 1 + cos d,
4(1 +cos0)p = 1. 2p = 3.
12. p = 4, 15. p = 2(l sin^),
pcos^=2. (l+sin^)p=l.
13. p = \/2, ^® P = cosd,
P = 2 sin 5. P = 1+ 2 cos d.
XIV, § 249] POLAR COORDINATES 391
Solve the following exercises by the use of polar coordinates :
17. Find the locus of the center of a circle which passes through a
fixed point O and has a radius 2.
18. Prove that if from any point a secant is drawn cutting a circle in
the points P and Q, then OP  OQ is constant for all positions of the
secant.
[Hint. By using equation (2), §246, show that the product of the
roots is constant.]
19. Secants are drawn to a circle through a fixed point on the cir
cumference. Find the locus of the middle points of their chord segments.
20. Find the locus of the middle points of the focal radii issuing from
one focus of an ellipse ; parabola ; hyperbola.
21. The focal radii of a parabola are produced a constant length. Find
the locus of their endpoints.
22. Through a fixed point on a fixed circle a variable secant OP is
drawn cutting the circle in B. If BP = 3 OB, find the locus of P.
CHAPTER XV
PARAMETRIC EQUATIONS
250. Parametric Equations. As a point P(xj y) moves along
a given curve, the x and y coordinates of the point vary. So
do many other quantities connected with this point, as for ex
ample, in general, its distance OP from the origin, the angle
which OP makes with the ajaxis, its distance from a fixed line,
etc. It is sometimes convenient to express x and y in terms of
one of these variables. This third variable, in terms of which
the variables x and y are expressed, is called a parameter. For
exaonple, we see that the coordinates of any point P{x, y) on
the circle whose center is at the origin and whose radius is r,
can be expressed in the form
(1) rx = rco
I 1/ = r si
r cos 0,
sin 6,
Fig. 230
where 6 is the angle XOP (Fig. 230). These are then para
metric equations of the circle. If we eliminate the parameter
6 between these two equations by squaring and adding them,
we obtain the equation 2 _j_ ,,2 _ ^2
X i y — r ,
which is the rectangular equation of the circle.
392
XV, § 250] PARAMETRIC EQUATIONS
Similarly, a pair of parametric equations of the ellipse
(2)
are
(3)
393
^4^ = 1
a2 62
f X = a cos 6,
1 y = 6 sin 6 ;
for, these values of x and y are seen to satisfy equation (2)
for all values of 6.
The geometric interpretation of equations (3) is important.
In Fig. 231, a geometric construction given in § 226 is used.
Fig. 231
The abscissa of P is equal to the abscissa of Q, i.e. a cos 0, the
ordinate of P is equal to the ordinate of R, i.e. b sin 6. There
fore the coordinates of P are x = a cos 0, y = b sin 0. The
angle is known as the eccentric angle of the point P. We
should note that is not the angle XOP.
A pair of parametric equations for the hyperbola
(4) ^^yl=i
^ ^ a^ b^
are
(5) x = a sec 9, y = b tan 9,
for these values of x and y satisfy the equation (4).
It is important to note that a given curve may have as many
sets of parametric equations as we please. For example, para
394 MATHEMATICAL ANALYSIS [XV, § 250
metric equations of a circle may be written in the form
x = a cos t, y = a sin t,
as above ; or they may be written in the form
or in any one of many other forms.
EXERCISES
1. Show that x = t, y =2 — t are parametric equations of a straight
line.
2. Show that x = I pt^, y =pt are a pair of parametric equations of
the parabola y'^ = 2 px.
3. Write two pairs of parametric equations for the line y = x.
4. Prove that ., ,„. „ ,
are parametric equations of a circle.
6. Write a pair of parametric equations for the rectangular hyperbola
a;2  ?/2 _ ^2,
6. Show that x = Acosd \ Bsin d, y = Asm 6 — Bcosd are para
metric equations of a circle.
7. Prove that x = 6 + 4 cos ^, y = — 2 + 3 sin ^ are parametric equa
tions of an ellipse.
8. Write a pair of parametric equations for the circle
(x  a)2 + (y  6)2 = r2.
9. Prove that a; = 6 + 4sec^, y=— 2 + 3 tan d are parametric equa
tions of a hyperbola.
10. Find the equation of the tangent to
X^ f/2
(<^) ^ + 7^ = ^» *^ ^1 = « <^08 ^i» yi = ft sin ^1
/v2 «f2
(ft) — o — t:; = 1' at ^1 = « sec 01, yi = 6 tan ^i.
(c) ?/2 = 2px, at xi = i ;)«i2, yi = pti.
XV, § 251]
PARAMETRIC EQUATIONS
395
11. Prove that the tangents to y'' = 2px at {^ph^.ph)^ ilPh^.Ph)
meet at the point [^^phti, \p{h + ^2)]
12. Write the equation of the tangent to y^= 4 ax at xi =
13. Show that
Wli^'
2a_
3a«
y =
3a«2
1 + «3 ' ^ 1+^3
are parametric equations of the curve x^ {y^ —Z axy.
14. Show that x = a cos^ e, y = a sin^ d are parametric equations of
21 2
the curve x'^ i y^ = a^.
15. Find the x and ^/equations of the curve whose parametric equa
tions are X = a (^ — sin d), y = a (1 — cos d).
251. Sketching Loci of Parametric Equations. If we assign
a series of values to the parameter and determine the series of
corresponding pairs of values for x and 2/j we can interpret
these values as the coordinates of points on a curve. Plotting
these points and sketching a curve through them, we have the
graph of the curve whose parametric equations were given.
Example. A pair of parametric equations giving the path of a
body projected horizontally from a height of 400 ft. with a velocity of
10 ft./sec., are x=10t, y = 400 — 16 t^. Sketch the locus.
t
1
2
3
4
5
X
10
20
30
40
50
y
400
884
336
256
144
Y
400
300
800
100
N
\
\
\
V
\
1
a
s
4
6
6
X
Fig. 232
In the preceding table are given the values of x and y corresponding
to the integral values of t from to 5 inclusive. Plotting these points
we have the graph in Fig. 232.
This curve is of course the same that we should obtain by first elimi
nating t and then plotting from the equation in x and y.
396 MATHEMATICAL ANALYSIS [XV, § 252
252. The Time as Parameter. Suppose a point moves in
a plane. At every instant of time t the point occupies a certain
position (x, y). In other words, the coordinates x and y of the
point P are functions of t, i.e.
,/,x r a; = a function of ty
I y = a function of t.
These equations are then parametric equations of the path
traversed by the point.
Such equations arise frequently in mechanics when it is
desired to describe the motion of a body subject to various
forces. For example, if a body is projected from a point
(0, 0) in a vertical plane at time i = 0, with an initial velocity
Vq, and making an angle a with the horizontal (icaxis), its
position at the end of t seconds * is given by the equations
{x = tv^ cos a,
[y = tVoSma}gt%
where, if Vq is measured in ft./sec, gr is a constant approxi
mately equal to 32.2. The use of these equations of a projectile
will be illustrated in the next article and the exercises follow
ing it.
EXERCISES
Sketch the following curves from their parametric equations.
1. x = t,
4. X = t'^  1,
7. x = t,
lo.
x = t,
2/ = « + 2.
2. x = r^,
y = t^\.
5. X = ^2 4. 1,
1
11.
y = ttK
X = sin d,
y = r.
y = z.
8. x = t\
y = cos d.
3. x = s+ 1,
6. x = t,
y = t\
12.
x = tan e,
y = s'^.
y\
9. x = t^+\,
y = t^^l.
y = sec d.
13. x = \Qt cos SO'',
14. x = 5«,
?/ = 25 + <
sin 30°  16 «2.
y = so
16 «2.
* The resistance of the air being neglected.
XV, § 253] PARAMETRIC EQUATIONS
397
253. The Path of a Projectile in Vacuo. The equations (7)
given in § 252 yield many results of interest regarding the paths of pro
jectiles. Some of these are given in this article and others are found in
the exercises below. They are, of course, only approximations to the
actual behavior of a projectile, in view of the fact that the resistance of
the air has been neglected *
By eliminating t between the equations (7), § 252, we obtain the equa
tion of the path in rectangular coordinates (x, y) :
(8)
y = X tan a —
gx^
Fig. 233
The path is, therefore, a parabola, with a vertical axis. The vertex
of the parabola is at the point (Fig. 233)
(9)
\m 2 a, <''''''' '
^9
The greatest height above the horizontal is
(10)
H =
The complete range, i.e. the distance from O to the point where the
projectile again meets the horizontal, is found as follows :
If in (7) we place y = 0, we find t = and
t = 2{vjfj) sin a.
The value of x for the second value of t found is the desired range B, i.e.
.2
B
^ sin 2 a.
g
This result could also have been found by placing y = in (8) and solv
ing for X. Why ?
* For the theoretical and practical discussion of the flight of actual projec
tiles (whose motion is 'appreciably affected by the resistance of the air) the
student is referred to Alger, The Groundwork of Naval Gunnery, or
External Ballistics.
398 MATHEMATICAL ANALYSIS [XV, § 253
EXERCISES
1. A gun is fired at an elevation of SO'^. Find the range if the muzzle
velocity of the shell is 1000 ft./sec. Aiis. About 5 mi.
2. What is the greatest height reached by the projectile in Ex. 1 ?
How long is its time of flight ?
3. What must be the initial velocity of a baseball thrown at an angle
of 45° in order that it may travel 200 ft. before hitting the ground ?
4. A stone is thrown from a tower 100 ft. high, with an angular ele
vation of 45° and an initial velocity of 64 ft./sec. How far from the foot
of the tower will the stone hit the ground ?
6. The great pyramid of Cheops is 450 ft. high. Its base is a square
746 ft. on a side. A ball is thrown upwards from the top in a direction
making an angle of 20° with the horizontal and with the velocity of
80 ft./sec. Will the ball clear the base of the pyramid ?
6. Prove that for a given initial angle of elevation the range of a pro
jectile is proportional to the square of the initial velocity.
7. Prove that for a given initial velocity the maximum range is
obtained when the angle of elevation is 45°.
8. Prove that with the notation of § 253, the time of flight of a pro
jectile from to (x, y) is (x/vo) sec a ; from Oto (J?, 0) is (2 Vo/g) sin a.
9. Prove that the paths of a projectile with given vq, but varying a,
have the same directrix.
10. Prove that the coordinates of the focus of the path of a projectile
are
( vq^ sin 2 a — ih)^ cos 2 a \
2sr ' 2g )'
Hence show that the locus of the foci of all paths in a given vertical plane
with the same vo is a circle with center at 0.
11. Prove that the parabola of maximum range has its focus on the
avaxis.
12. Prove that the locus of the vertices of the paths with given "yo is
an arc of an ellipse.
13. Prove that the locus of the vertices of the paths with a given a.
and a varying vo is a straight line.
14. Prove that the locus of the foci of the paths with a given a and a
varying vo is a straight line.
XV, § 254j
PARAMETRIC EQUATIONS
399
H
^
254. Locus Problems. Parametric equations of a curve
are sometimes much more easily obtained and easier to work
with than either the rectangular or polar equations. The
following problems illustrate some
of the methods that may be em
ployed.
Example 1. A line of fixed length
moves so that its ends always remain on
tlie coordinate axes. Find the locus gen
erated by any point of the line.
Call the point whose locus is desired
P(x, y). Since the line is of fixed length,
call the segments into which P divides it,
x = a cosd^ y=b sin d.
Example 2.
a fixed line.
Take for origin the point O where the moving point P touches the
fixed line. If r is the radius of the circle and the angle PCD (Fig. 235)
is 6 radians, then PD = rsind, DC = r cos 6 and OB = arc BP = rd.
Fig. 234
a and b (Fig. 234). Then
Therefore the locus of P is an ellipse (§ 250).
Find the locus of a point P on a circle which rolls along
Fig. 235
Now if P is denoted by the coordinates (aj, y),
x= 0A= OB AB= OB PD = rd r sine = r(d sin d),
y = APz=BC DC = rr cos d = r{l cos d).
Therefore
ni\ {x = r(6— sin^),
\y = r(l — cos d)
are parametric equations of the curve traced by the point P. This curve
is known as the cycloid.
400
MATHEMATICAL ANALYSIS [XV, § 254
Example 3. Find the locus of a point P on a circle of radius a which
rolls on the inside of a circle of radius 4 a.
Take the center of the fixed circle as the origin and let the xaxis pass
through a point M where the moving point P touches the large circle.
Let angle MOB = d radians. Now
we have arc PB = arc MB = 4 ad
and arc PB = a x angle PCB.
Therefore
a X angle PCB = 4 ad,
or angle PCB = 4 ^.
But
Z OCD + ZDCP^Z PCB = T.
Therefore
0 + Z2)CP+4d = T,
i.e. ZDCP=Se.
Fig. 236
Now if the point P is denoted by
{x, y) we have
x = OE=OD\ DE = OD\ FP = OCcose + CP&mlZ e\
= 3 a cos ^ + a cos 3 ^ = 4 a cos* ^,*
y = EP=DG'FC= OC sine  CPcoslSe\
= 3 a sin ^ — a sin 3 ^ = 4 a sin* ^ ;♦
that is,
(12) a: = 4acos3^, y = 4a8in^0.
This curve is called the fourcusped hypocycloid.
Example 4. P8P' is a double ordinate of an ellipse ; Q is any point
on the curve. If QP, QP' meet the xaxis in O and 0', respectively,
prove that CO • CO' = a^, where C is the center of the ellipse.
Let P be (a cos ^i, ft sin di), then P' is (acos^i, — ftsin^i). Let Q
be (a cos 6, b sin d). The equation of line PQ is
y " sin a = —^ (X — a cos a)
a(cos^i — cos^)
* Prove that cos 3 ^ = cos (2 + ^) = 4 coss — 3 cos ^,
sin 30 = sin (2 + ^) = 3sin0— 4 8in«tf.
XV, § 254] PARAMETRIC EQUATIONS 401
and its x intercept {i.e. CO) is
a(cos 6 sin d^ — sin 6 cos 6i)
sin d — sin di
Similarly CO' is «(  cos ^ sin ^i  sin cos gQ ,
sin ^ + sin di
The product CO • CO' gives a"^.
EXERCISES
1. A line of fixed length 2 a moves with its ends always remaining on
the coordinate axes. Find the locus of the midpoint of the line.
2. Find the locus of the middle points of chords of an ellipse drawn
through the positive end of the minor axis.
3. Find the locus of a point P' on the radius CP of the cycloid (Fig.
235) if CP' = b and b<r.
4. The same as Ex. 3, except & > r.
6. A circle of radius r rolls on the inside of a circle of radius a.
Find the locus of a point P on the moving circle.
a — r
Ans. The hypocycloid
X = (a — r) cos d { r cos
r
a — r
= (a — r) sin ^ — r sin —  — <
Ans. The epicycloid
X = (a \ r) cos 6 — r cos
y = {a h r) sin ^ — r sin
where 6 is the same as in Example 3, § 254.
6. A circle of radius r rolls on the outside of a circle of radius a.
Find the locus of a point P on the moving circle.
a + r
a + r
—re,
where 6 is the same as in in Example 3, § 254.
7. The area of the triangle inscribed in an ellipse, if ^i, ^2* ^3 are the
eccentric angles of the vertices, is
\ ah [sin (^2 — ^3) + sin (^3 — ^1) + sin (^1 — ^2)]
^2 ah sin ^2ii_^ sin h=Jl sin ^JZLb,
2 2 2
8. The coordinates of one extremity of a diameter of an ellipse are
(a cos ^1, ft sin ^1). Show that the coordinates of one extremity of the
conjugate diameter are (— a sin ^1, h cos ^1).
2d
PART IV. GENERAL ALGEBRAIC METHODS
THE GENERAL POLYNOMIAL FUNCTION
CHAPTER XVI
MISCELLANEOUS ALGEBRAIC METHODS
255. The Need of other Methods. We have hitherto con
sidered special functions such as a;2, sin ic, log^QX, or special
types of functions such as mx { h, ax^ + bx {c, log„ x, a'' ; and
we have studied their geometric and other applications. The
study of more general types of functions, for example, the
general polynomial of degree n,
a^x"" + a^_iX'^~^ h h «i^ + «o)
of which mx + h, ax^ f 6a3 f c, ay? + bx^ \ cx^ d are special
types, requires more powerful methods. Some of these we
propose to consider in the present and the succeeding chapters*
256. Technique of Polynomials. We shall first recall the
technique of the addition and multiplication of polynomials.
We begin by noting that a polynomial
can be completely represented by so called detached coeffi
cients, as follows : « ^ ^ «
the place of each coefficient in this expression indicating the
power of X to which it belongs. Thus, for example,
2316
402
XVI, § 256] ALGEBRAIC METHODS 403
represents the polynomial 2oi:^ — Sx^ + x\6; and
10 01
represents the polynominal x^ — 1.*
To add two or more polynomials we need merely add the
coefficients of like powers of x. Thus, the sum of x"^ — l,x*\
2s(r^\4:X^ + Sx\5, and 2a^ — 5x'^\x\lis given by
101
12 4 3 5
25 1 1
1 4 4 5=a;* + 4a^f4a; + 5.
The analogy of this process with that of adding a column of
numbers may be noted.
The product of two polynomials A and B is obtained by
multiplying A by each term of B and adding the results.
Why ? The multiplication of two polynomials by the method
of detached coefficients is also quite analogous to the familiar
method of multiplying two integers. Thus the product of
2 a^ \ Sx"^ — x —2 by aj2 + a; + 4 is given by
23 1 2x114
2 312
2 312
8 1248
2 5 10 96 8=2a^\5x'\10a^\9x^6xS.
The student should convince himself, by multiplying these
polynomials in the ordinary way, that the above method is
indeed valid.
♦This method of representing polynomials will seem very natural, if we
note the analogy with the familiar method of representing integers. The
number 217 is simply a short way of writing 2 x 102 + 1 x 10 + 7, i.e. the value
of the polynomial 2z^{x + 7, when x = 10. We have, therefore, been famil
iar with the method of detached coefficients from the time when we first
learned to write numbers.
404 MATHEMATICAL ANALYSIS [XYI, § 257
257. The Division Transformation. A polynomial B is
said to be a factor of a polynomial A, if there exists a poly
nomial Q such that A = BQ. A is then said to be divisible
by B. If no such polynomial Q exists, and if the degree of B
is less than that of A, we may always determine polynomials
Q and E, such that
(1) A = BQ\E.
Furthermore, the remainder R can always be so determined
that its degree is less than the degree of B.
The process whereby a given polynominal A is expressed
in terms of another polynomial B in the form (1), i.e. the
process of finding Q and R, when A and B are given, is
called the division transformation. That it is always pos
sible to find polynomials Q and R, if the degree of B is
less than that of A, will be clear from the consideration of
the following example.
Example. Given A = 2x* + 5x^— 'Jx'^ + 12x — 5 and B = x^10x
+ 8 to find a polynomial Q such that ^ — J5^ is a polynomial of degree
less than that of B.
Since the term of highest degree in J. is 2 x* and that in B is x^, it ap
pears that A — 2 x^B can contain no term of degree higher than 3. In
fact, we find A — 2 x^B = 25 x^ — 23 x^ f 12 x — 5. Similarly, since the
term of highest degree in ^ — 2 x^B is 25 x^, we see that the expression
{A — 2x^B)— 25 xB can contain no term of degree higher than 2. By
continuing this process we shall arrive at a polynomial which is of degree
less than that of B. The work may be arranged as follows.
^=2x* + 6 x8 7x21 12x
J5. 2x2 = 2x4 20x3 + 16x2
x210x + 8 = 5
2 x2 + 25 X 4 227
^5.2x2 =
J525x =
25x8 23x2} 12X6
25x8250x2+ 200 X
^^(2x2 + 25x) =
J? . 227 =
?(2x2 425x + 227) =
227x2 188X5
227x22270x+1816
2082x1821 = 7
XVI, § 258] ALGEBRAIC METHODS 405
If the meaning of each step in the process is followed by means of the
expressions written at the left,* it will be seen that the process has deter
mined a polynomial ^ = 2^^ + 23x + 227
such that A— BQ = B,
where B is of degree less than that of B. The nature of the process
shows that finally such a polynomial B will always be reached.
258. Remarks on the Division Transformation. While
the process discussed in the last article is known as the
division transformatioyi, it is not a process of division. Only
if we take a further step and divide both members of the
identity (1) (§ 257) by B, to obtain
(2) =0 + ,
do we really divide A by B. The importance of this distinction
lies in the fact that the relation (1) as derived above is valid
without distinction for all values of the variable x involved.
For in deriving the relation we made use only of the opera
tions of multiplication and subtraction. However, the relation
(2) becomes meaningless for all values of x for which 5 = 0.
We assumed in deriving the relation (1) that the degree
of B was less than that of A. This is indeed necessary if Q is
to be a proper polynomial. However, if the degree of B is
equal to that of A, the same process will lead to a relation
(3) A = B'q\R,
where g is a constant. If the degree of B is greater than that
of A, we may obviously write the trivial relation
(4) A=B'()^A
where A equals R and is by hypothesis of lower degree than B.
If we consider a constant as a poljoiomial of degree 0, the last
* These expressions are not of course a necessary part of the process.
They are given here only to facilitate understanding.
406 ' MATHEMATICAL ANALYSIS [XVI, § 258
two cases may be included in the form (1). Our theorem then
takes the general form
Given any two polynomials A and B of degrees greater than 0,
the7i polynomials Q and R can always he found such that for all
values of the variable we have A = BQ { R where R is either
zero or of degree less than that of B.
Moreover, the transformation of A to the form BQ\ R is
unique, i.e. there exist just one polynomial Q and one poly
nomial R satisfying the conditions of the theorem.
For, suppose there were a second pair, for example Q' and
R'. We should then have BQ+ R = B'Q' + R', or B{Q  Q')
= R' — R. But R' — R is either zero or of degree less than
that of B, while B(Q— Q') is either zero or degree equal to
or greater than that of B. Hence both are equal to zero and
R = R', Q=Q'.
EXERCISES
1. Add the following polynomials by means of detached coefficients :
(a) 2ic2 + 7x + 1, 5a;2 + 2, 3x3 + 4x8.
(6) 6 «2 + 5 ( + 1, 9 <2 + 8 « + 3, 6 «3 + 2 « + 1.
(c) ay^ \ by + c, 2 ay^ + 3by + i, 3 ay^ + 6by + 7 c.
(d) 4 a2 f 3 a + 2, 6 aM 1, 4 a2 + 2 a + 3, 2 a2 + 6 a.
2. Perform the following multiplications by means of detached coeffi
cients :
(a) «3 + 2x2 + X + 3 by 2 X + 1. (c) a* + 1 by a'^ + 1.
(b) x8 + 3x2 + 4 by x3 + X + 2. {d) y^ + 1 y + \2\)y y^ + 3y + 2.
3. In each of the cases below transform A into the form BQ >r B,
where B is of degree less than B. Also write down the corresponding
form for A/B. Detached coefficients may be used to advantage.
(a) ^ = 6x4 4 7 x3  3 x2  24 X  20, B = 3 x2 + X  6.
(6) ^=3x4 + 2x332x266x35, £ = x22x7.
(c) J[ = 2 x6 + 5 x8 + 13 x2 + 2 X, 5 = x2 f 2 X + 4.
(d) ^ = 4x7^3x^ 19x* + 2x34x24x47, B = x8x6.
XVI, § 259] ALGEBRAIC METHODS 407
4. Determine m and n so that k* — moc^ ^x"^  nx + \ may be exactly
divisible by a;2 + 2 X + 1
6. Prove the following propositions :
(a) If we multiply the dividend A by any constant as k{k ^ 0), we
multiply the quotient and the remainder by k.
(6) If we multiply the divisor by A; (A; ^t 0), we divide the quotient by
k but leave the remainder unchanged.
(c) If we multiply both dividend and divisor by k{k ^ 0), we multiply
the remainder by k but leave the quotient unchanged.
259. The Highest Common Factor of two Polynomials.
Two polynomials A and B may or may not have a common
factor of degree greater than 0, i.e. a polynomial F (of degree
greater than 0) may or may not exist such that A = FQ^
B = FQ' where Q and Q' are also polynomials. If no such
polynomial F of degree greater than exists, then A and B
are said to be prime to each other. If, on the other hand,
they have a common factor of degree greater than 0, the one of
the highest degree is called the highest common factor (H. C. F.).
Theorem 1. If A and B are polynomials with a common
factor and M and N are any two polyyiomials, then any common
factor of A and B is a factor of AM\ BN.
For let F be any common factor of A and B. Then we
have A=::FQ and B=FQ'. Therefore
AM\ BN= F(QM+ Q'N),
which shows that i^ is a factor of AM + BN.
Theorem 2. If A, B, Q, R are polyyiomials such that
A = BQ\ B, the common factors of A and B are the same
as the common factors of B and R.
For, by Theorem 1, any common factor of B and i2 is a
factor of A and hence a common factor of A and B. More
over, from the relation A — BQ = R and the last theorem, any
common factor of A and B is a fac or of B and R.
408 MATHEMATICAL ANALYSIS [XVL § 259
Successive applications of the division transformation there
fore enable us to find the H. C. F. of two polynomials A and B
as follows :
Using the division transformation on A and B, we may
™*^ A = BQ + B,
where the degree of R is less than that of B. li B = 0, B is
the H. C. F. If 72 is a constant different from zero, then A
and B have no H. C. F. Why ? If the degree of B is at least
equal to 1, we may use the division transformation on B and B
to obtain B = BQ,hB„
where Bi is of degree less than B. If Bi = then B is the
H. C. F. of ^ and B. If Bi is a constant different from zero,
A and B are prime to each other. If B^ is of degree at least
equal to 1, proceed as before, expressing B in the form
B = B1Q2 { B2.
This process may be continued until a remainder B^ is
reached which is either zero, or a constant different from zero.
If B^ = 0, then B,^_i is the H. C. F. sought. If B^ is a constant
different from zero, then A and B are prime to each other.
Example 1. Find the H.C. F. of 4a:8_3a;2+7a;l and 2 a;23a;l.
A = ^x^ Sx'^ + I x1
4a;86x2 + 2 x
2x^3x + l = B
2x + ^ = Q
3 x2 + 6 x1
3x2 f a; + 
Replace Bhj x — ^% — E'.
B = 2x^3 x + l \xj^g=B
2x:2\%x \ 2x\i
nx + VA
Therefore A and B are prime to each other.
XVI, § 260]
ALGEBRAIC METHODS
409
Example 2. Find the H. C. F. of x^  2 x2 f x + 4 and 3 x^ f 8 x2 +
3x2.
The work may be arranged as follows. Since the H. C. F. of two
polynomials is not altered in any essential way by multiplying or dividing
either of them by any constant (^^ 0), we can avoid fractional coefficients.
X3
X3
2x2
+
xf4
X
2x2
2x2
+
2x + 4
42
21
2x + 2
X41
3x3
4
8x2
4 3
X 2
3x3
—
6x^
+ 3
x + 12
14x2
14
14
X2
1
X2
+ x
— X
1
— X
1
Therefore the H. C. F. is x 4 1
EXERCISES
Find the H. C. F. of each of the following pairs of expressions :
(a)
(&)
(c)
(d)
(/)
x3 42x213x+ 10, x3 + x2 10x48.
3x* 4 5x2 + 2, x6  4 X* 4 5 x2  2.
a;8_2x222x + 8, x2 6x4 2.
a3 + 3 a2  3 a  5, aa
a'
4 3 a 4 5.
y^ — yf^ — y + i? y'^ \ y + i
&4 + 5.3 ^ 6 ^2 _^ 5 5 ^. 5^ ^5 4. 4 53 _^ 52 _
1 4 xx2 5x3 + 4xS 14x843x4.
6 6 41.
{h) 4x45x3 + x + 1,3x44x3+1.
(i) a^ 4 a3 + a 4 1, a2 + a + 1.
(j) x^>l,xl.
2. Prove that, if the coefficients of two polynomials are rational (or
real), the coefficients of the H. C. F. are rational (or real).
3. If F is a factor of A but not of B, how does the H. C. F. of ^4 and
FB compare with the H. C. F. of A and ^ ? In introducing and suppress
ing factors during the process of division, what precaution must be taken
and why ?
260. Functional Notation. We have already used special
notations to represent special functions. Thus sin, cos, tan,
log, etc. are special notations with which we are familiar. We
shall now introduce a notation that is more general, for it is
410 MATHEMATICAL ANALYSIS [XVI, § 260
applicable to all kinds of functions. We shall use the symbols
f{x)j F(x), (t>{x), — to represent various functions of x, and
can then speak of the "/function," " (^function," etc., just as
we speak of the sine function, logarithmic function, etc.
Moreover, such a notation can be used to represent the value
of the function in question for a given value of the variable.
Thus, if f(x) is used to represent the function x^ { 2x — 1, the
symbol /(2) denotes the value of this function when x = 2
(just as sin (7r/2) means the value of sin x, when x = 7r/2) ; i.e.
/(2) =22 + 2. 21 = 7.
Similarly, with the meaning just given tof(x), we should have
f(x }h) = {x + hy ^2ix + h) 1.
It should be noted that, when a certain function is called
f(x), then, throughout any discussion where this function is
used, f(x) always means that particular function and no other.
EXERCISES
1. Given/(x) = 3x2 + 2a; 4, fiiid/(2),/(^),/(0),/(a; + 1/x).
2. Given 0(x) = x/(x  1), find 0(2), (f>(x + h), 0(1  x), 0(10).
3. Given F{x) = e* + e^ find F(0), F(l).
4. Given/(x) = (x  l)/(a; + 1), prove that
6. If 0(x + y) = (t>{x) + 0(j/), show that 0(3 x) = 3 0(a;).
6. Given f(x) = 2 x Vl — x^, prove that
/(sinx) =/(cosx) =sin2x.
1
X
Given /(x) = , find the value of /Ml+^V
X
8. Given ^(x) = e* + e"*, prove that
e{x^y)d{xy) =^(2x) }^(2y).
9. Express tlie fact that the volume of a sphere is a function of its radius.
XVI, § 262] ALGEBRAIC METHODS 411
10. Given F {x) = (x \/x){x'^  \/x'^){xJ^  \/x?), prove that
F{z) = F{\/z),
11. Given that/(w) = n !,* prove that (n + 1) f{n) =f(n + 1).
12. Given that/(.7;)=a;2f2, find/[/(x)].
13. Given /(x) = sinx, find/(7r/2),/(7r/3), /(tt).
14. Given /(x) = sin x and <f)(x) = cos x, prove that
(«) lf(x)T + [0(^)]^  1. (d) fix) = «/,(7r/2  X).
(b) f(x)  0(a) = tan X. (e) /( x) = /(x).
(c) /(.r + 2/) =f(x) cf>(y) +f(y) <p{x). (/) 0(x) = 0(  x).
16. If /(x) = loga X, show that
(a) / (a;) + / (y) =f{xy). (c) /(xA) + f(y/x) = 0.
(6) /(x") = n/(x).
16. What functions may/(x) represent when
(a) fix + y) =f(ix)f(y). (c) /(x") = n/(x).
(6) /(x+ 2/) =/(x) +/(2/). (d) /(^^ =/(2/) /(x).
261. The Remainder Theorem. If a polynomial f(x) is
divided by x— a, the remainder isf(a).
If f(x) is the dividend, a; — a is the divisor, Q(x) the
quotient, and R the remainder, then
(5) f{x)=(xa)Q(x)^E
where B, since it is of lower degree than x — a, does not in
volve X at all, i. e. E has the same value for all values of x.
Since the values of the two members of this identity are
equal to each other for all values of x, we have for the par
ticular value X = a. ^/ s r>
J {a) = M.
262. Factor Theorem. If f(x) is a polynomial and a is a
number such that f (a) = 0, then x — a is a factor off(x).
The proof of this theorem is left as an exercise.
*n! = l2.3.4 •... • w, that is, 2! =2, 3! =6, 4! =24, ....
412 MATHEMATICAL ANALYSIS [XVI, § 262
EXERCISES
By use of the remainder theorem find the remainder when
1. x^ — 2 x'^ + 3 is divided by a; — 2.
2. a:i3 _ 45 a:i2 + 26 x^ + 12 is divided by a;  1.
3. a:i2 + 1 is divided by jc + l ; by a; — 1.
4. Show that — 2 is a root of the equation 2 x^ + 3 a;^ — 4a; — 4 = 0.
5. Show that a;'* + «" is divisible by x + a if w is odd.
6. Show that x" + a" is not divisible by x + a if n is even.
7. By means of the remainder theorem find k so that x^\ kx^\2x +^
may be exactly divisible by a; — 2.
8. Find the polynomial in x of the second degree which vanishes when
x= I and when a; = 2, and which assumes the value 10 when x = 3.
263. Synthetic Division. We shall proceed to explain a
simple method of performing the division transformation
when the divisor has the form x — k, i.e. when the divisor is a
binomial of the first degree in which the leading coefficient is 1.
Let the given polynomial be
a„aj" + «„iaJ"~^ + a^2X"'^ + ••• + OjX + a^
and the divisor x — k.
The ordinary process of long division leads to
X — k
a,.a;^— a„ A;.r"i ^«^"~^ + (^«n + ^ni)^""^
(ka, + ft^ Qa;"^  k(ka^ + a.,i)x^^
It is now not difficult to see that
(a) the first coefficient in the quotient is a„, i.e. the coeffi
cient of the leading term in the dividend ;
(b) the second coefficient in the quotient is obtained by
multiplying the first coefficient of the quotient by k and adding
to it the second coefficient of the dividend :
XVI, § 263] ALGEBRAIC METHODS 413
(c) the third coefficient of the quotient is obtained by mul
tiplying the second coefficient of the quotient by k and adding
to it the second coefficient of the dividend.
We may then arrange the work as follows :
a„ a„_i a„_2 «„3  1^
or„ ka^ f a„_i k{ka„ + a„_i) + a„_2 
Here the first line contains the coefficients of the dividend
in order and the third line gives the coefficients of the quotient
and the remainder in order. Every number in the third line,
after the first, is obtained by multiplying the preceding num
ber by k and adding to it the next number in the first line.
Example 1. By synthetic division, find the quotient and the re
mainder when X*— 2ic3fx2 + 3a;— 2is divided by cc + 2.
Solution : 121 3—21 — 2
 2 8  18 30
1 _ 4 9  15 28
Hence the quotient is x^ — 4 ic^ + 9 x — 15 and the remainder is 28.
Example 2. If/(x) = 2x* +3x3+7x2} 14x + 20, find/(3).
2 3 7 14 2013
6 9 48 102
2 3 16 34 122=/(3). Ans.
EXERCISES
In the following exercises use synthetic division :
1. Find the remainder when x^ + 3 x^ — 6 x + 2 is divided by x — 2.
2. Find the remainder and the quotient when x* — 3x'^ + 2x + 3 is
divided by x + 3.
3. Show that 3 is a root of the equation x^  4 x^  17 x + 60 = 0.
4. Find k so that 3 is a root of the equation x'*  3 x^ + A:x + 1 = 0.
5. Is 5 a root of the equation x^ — x^ + 7 = ?
414 MATHEMATICAL ANALYSIS [XVI, § 264
264. Properties of Integers. We have already noticed
(ftn.j p. 403) that the familiar method of writing an integer
greater than 9 represents it as the value of a polynomial.
Integers and polynomials, therefore, have many properties in
common. We may, for example, gain an insight into the reason
for the rules of arithmetic used in adding a column of figures
or in finding the product of two integers by comparing these
rules with the technique of adding and multiplying polynomials
discussed in § 256 *. We shall proceed to discuss some of the
properties of integers relating to divisibility, etc., which are
valuable in our everyday use of numbers.
265. Prime and Composite Numbers. An integer greater
than 1 that is divisible by no integer except itself and 1 is called a prime
number or simply a prime. Thus 2, 3, 5, 7, 13 are prime numbers. Any
integer (> 1) not a prime is called a composite number. Any composite
number is the product of two or more primes, thus 6 = 23, 100 = 2 • 2 •
5 . 5 = 22 . 5^. Any composite number 7i may be resolved into its prime
factors in one and only one way. When resolved it has the form
n = j?i*ij92"* •••Pfc"* When a number has been resolved into its prime
factors any question regarding its divisibility is readily answered by the
following theorem.
A number a is divisible by a number b if and only if every prime factor
of b occurs in a to at least as high a power as it occurs in b. This
theorem follows readily from exercises 15 and 17 below. The proof is
left to the student. As an illustration, if a = 2 • 3^ • 17^ • 37 and 6 =
2 • 32 . 17 we recognize at once that a is divisible by b and that the quo
tient is 3 • 17 • 37. If, on the other hand, b were 2^ . 3^ • 17 then a would
not be divisible by &.
The common factors of two integers are also readily found if the num
bers have been resolved into their prime factors. Why ? Two integers
which have no common factor (> 1) are said to be prime to each other.
The notion of prime numbers and the investigation of their properties
is very ancient and to this day some of the most difficult problems of
advanced mathematics relate to this field. Some of the properties are
quite elementary, however, and have been listed below in exercises.
* Carrying this comparison out in detail forms a valuable exercise. The
familiar process of "carrying" a digit from one column to the next is about
the only thing that requires special investigation.
XVI, § 265] ALGEBRAIC METHODS 415
EXERCISES
1. Prove that if a and h are both divisible by w, then a\h and a—b
are divisible by n and a • 6 is divisible by n^. Is a similar theorem true
of polynomials ?
2. Prove that a product of any number of integers is divisible by n if
one of the integers is divisible by n. Is a similar theorem true of poly
nomials ?
3. If c = a & is divisible by «, must either a or 6 be divisible by n ?
4. Prove that if a number a leaves a remainder r when divided by &,
then the number obtained by adding to a any multiple of h will leave the
same remainder.
5. Prove that if the last digit on the right of an integer is even, the
integer is divisible by 2.
6. Prove that if the number formed by the last two digits of an
integer is divisible by 4, then the number is divisible by 4.
7. Prove that if the number formed by the last three digits of an
integer is divisible by 8, then the integer is divisible by 8.
8. Prove that if the last digit of an integer is or 5 then the integer
is divisible by 5.
9. Prove that if the sum of the digits of an integer is divisible by 3
(or 9) then the integer is divisible by 3 (or 9).
10. Prove that if the sum of the first, third, fifth, etc. digits of an
integer is equal to the sum of the second, fourth, etc., the number is
divisible by 11.
11. If the sum of the digits of an even number is divisible by 3, the
number itself is divisible by 6.
12. Determine without performing the division whether the following
numbers are divisible by 2, 3, 4, 5, 6, 8, 9, 11.
(a) 2562. (c) 123453. (e) 127898712. {g) 111111111111.
(6) 12345. {d) 2673. (/) 7325 x 8931. Qi) 11111111112.
13. How would you recognize that a number is divisible by 45 ?
14. Prove that if the product « • & is divisible by a prime number p,
either a is divisible by p or 6 is divisible by p. Is a similar theorem true
for polynomials ?
15. Prove that if a number c is a factor of ah and is prime to a, it
must be a factor of h. Is a similar theorem true for polynomials ?
16. Prove that the quotients of two numbers by their H. C. F. are two
numbers prime to each other. Is a similar theorem true for polynomials ?
416 MATHEMATICAL ANALYSIS XVI, § 265
17. Show that if a number is divisible by each of two numbers which
are prime to each other, it is divisible by their product. Is a similar theo
rem true for polynomials ?
18. Show that the product of two numbers is equal to the product of
their H. C. F. by their L. C. M. Is a similar theorem true of polynomials ?
19. Prove that the number of primes is unlimited.
[Hint. Suppose that the theorem were not true and that p were the
greatest prime. Let pi, /)2, P3, •••, Pni be the set of all primes less than p
and consider the number
PiP'zPz ■■' PniP^ 1
This number is not divisible by any of the primes pi, p2, •••, p. The rest
of the proof should be obvious. This proof was first given by Euclid.]
20. By trying successive primes 2, 3, 5, 7, •••, determine whetlier or not
1009 and 1007 are primes. In this case we may stop with the prime 31.
Wiiy ? Ans. 1009 is prime.
• 21. Resolve into prime factors the numbers 604800 and 12250.
22. Is the number 2^31253 a perfect square ? Is it a perfect cube ?
23. Show that the relation ah — cd = 1, where a, &, c, rf, represent in
tegers, is not possible unless a and c are prime to each other.
24. Two consecutive integers are always prime to each other. Is this
true also of any two numbers differing by 7 ?
25. What is the smallest cube of masonry that can be constructed of
bricks 8x3x2 inches ? It is assumed that the bricks are placed so that
any two equal sides are parallel.
266. Partial Fractions, in certain problems it is sometimes found
necessary to express a fraction in which the numerator and denominator
are polynomials in one variable as the sum of several fractions each of
which has a linear or at most a quadratic function in the denominator.
In what follows it will always be assumed that the given fraction is a
proper fraction^ i.e. a fraction in which the degree of the numerator is less
than that of the denominator. Any fraction which is not proper can be
written as the sum of a polynomial and a proper fraction. Therefore our
problem may be stated as follows : To express a proper fraction as the
sum of several proper fractions.
The method of approach is to assume that the fraction can be expressed
in the desired form and then seek to determine the numerators which in
the assumed form are left undetermined. Four distinct cases arise.
XVI, § 266] ALGEBRAIC METHODS 417
Case I. When the denominator can be resolved into factors of the first
degree all of which are real and distinct.
Example 1. Resolve into partial fractions
9x^ a;2
x^ — X
The sum of the three fractions
X X —\ X \l
will give a fraction whose denominator is x^ — x. We therefore try tc
determine A^ B, C so that
(Q) 9x^x2 _A B C
X^ — X X X — 1 x \ 1
Clearing of fractions we have
(7) 9x2x2 = A{x^  1) + B{x^ ■{■ x) { C {x'^  x) .
Since (7) is to be true for all values of x, we seek values of A, B, O, such
that the coefficients of like powers of x will be equal, i.e. such that
A + B\ C^9, BC = l, A =2.
Solving these equations, we find A = 2, B = S, (7 = 4. Hence
9a;2a;2_2 3 , 4
X^ — X x X—1 X \ I
Case II. When the denominator can be resolved into real linear fac
tors some of which are repeated.
Example 2. Resolve into partial fractions
2 a;2  X + 2
x(xiy
The sum of the fractions
A^ B , C
X x\ {X 1)2
will give a fraction whose denominator is x{x — l)^. Therefore we shall
try to determine A, B, C so that
2x^x + 2^A^ B , C
X{X  1)2 X Xl {X 1)2
Clearing of fractions, we have
2x2342 = A{x 1)2 J^ Bx{x 1)+ Cx.
Equating the coefficients of like powers of x, we have
A^B = 2, 2AB\C = \, A = 2.
2e
418 MATHEMATICAL ANALYSIS [XVI, § 266
Solving for A, B, O, we find ^ = 2, B = 0, C = 3. Hence
2x2a;+2_2 3
x(a'l)2 X, (a;l)2
The assumptions to be made under Cases I and II are contained in the
following rules.
(1) For every unrepealed factor x— a of the denominator^ assume a
fraction of the form A/{x — a).
(2) For every repeated factor {x — a)* of the denominator^ assume a
sum of fractions of the form
A\ ^ A2 _^ ,,, I Ak
X — a (x— a)2 (x — ay
Case III. W?ien the denominator contains quadratic factors which
are not repeated.
Example 3. Resolve into partial fractions
5 x2 + 4 X + 3
(X+1)(X2 + 1)'
Let 5 x2 + 4 X + 3 ^ A Bx + G
(x+ l)(x2+ 1) x + 1 x^+ 1 '
Clearing of fractions, we have
5 a;2 + 4 X + 3 = A(x^ +1) {(Bx+ C)(x + 1)
= Ax^ + A { Bx'^ ^ Bx + Cx + a
Equating coefficients,
A^B=o, B+C = 4,A+C=S.
Therefore A = 2, B = S, 0=1. Hence
5x^ + 4x+3 ^ 2 . 3x41
(x+l)(x2 + l) x+1 X241'
Case IV. When the denominator contains quadratic factors which
are repeated.
Example. Resolve into partial fractions
3 x^ + x'^ + 8 x'! + X + 2
X(X2 + 1)2
Let 3 x^ + x^ + 8 x2 + a; + 2 ^A Bx + C Dx + E
x(x2 +1)2 X x2 + 1 (x2 f 1)^'
Then,
3 x* + x8 + 8 x2 + X + 2 = .4(x2 + l)a + (^x + C)x(x2 + 1) + {Dx + E)x
= ^x* + 2 ^x2 + A + Bx^ + Cx3 + ^x2 + Cx + Z)x2 + Ex.
XVI, § 266] ALGEBRAIC METHODS 419
Equating coefi&cients we have
^ + JB = 3, C=l, 2^ + 5 + i) = 8, C+E = l, A = 2.
Hence, A = 2, B = 1, 0=1, D = 3, E = 0.
Therefore, ^^^ ^ x^ + Sx'^ + x + 2 ^2 x±l_ Sx
x(x^+iy^ X a:2 f 1 (a;2 + 1)2
The assumptions to be made in Cases III and IV are contained in the
following rules.
3. Corresponding to every unrepealed factor of the form ax'^ + hx \ c,
assume the partial fraction Ax A B
ax? + 5x + c
4. Corresponding to evet^ factor (ax^ \ bx + c)*, assume the sum
Aix + Bi J A2X + B2 __ ^^^ AkX 4 Bk
ax^ ^ bx \ c (aa;2 f 6x + c)2 (q[x2 + 6a: + c)*'
EXERCISES
Resolve into partial fractions each of the following fractions.
J 4x+ 1 jj 3 a:2  5 X + 4
(a;_l)(x+l)(x + 3)
3a;l
x24 '
2x + l
x2(x — 4)
x^ + 1
X(X 1)8*
1
x%x + 1) *
x2 + 2 a; + 1
x^ + x
2 x2  1
3x3 + 3 x'
2x + l
X3 + X2 + X *
1
12.
13.
14.
16.
16.
17.
18.
19
X(X2 + 1)2
3
X81' '" X2(X2+1)2
10. ^. 20
(X  1)3
X2
(X21)2'
X*
(x2l)(x + 2)
x2
(X + 1)X2
2 x2  X + 3
x(x2l)(2x3)
1 +x^
(2  x2) (2 + x2) '
X* + X2
X* + X2 + 1
32x2
(2  3 X + x2)2
5 x3 4 2 X + 1
(X^+1)(X1)2°
2x+l
CHAPTER XVII
PERMUTATIONS, COMBINATIONS, AND PROBABILITY
THE BINOMIAL THEOREM
267. Definitions. Suppose that a group of n objects is
given. Any set of r (r ^ n) of these objects, considered with
out regard to order, is called a combination of the n objects
taJce7i r at a time. We often denote the objects in question,
which may be of any kind, by letters, as a, 6, c, •••, k. The
number of combinations of these n letters taken r at a time is
denoted by the symbol „C^. For example, the combinations
two at a time of the four letters a, b, c, d are,
ab, ac, ad, be, bd, cd.
Since there are 6 of these combinations in all, we have 4C2 = 6.
On the other hand, any arrangement of r of these n objects
in a definite order in a row is called a permutation of the n
objects taken r at a time. The symbol „P^ is used to denote
the number of such permutations.
For example, the permutations of the four letters a, b, c, d
taken two at a time are
ab ac ad be bd cd ba ea da cb db dc.
Since there are 12 of these arrangements in all we have
4P2 = 12. We have assumed in these examples that the
objects are all different, and that the repetition of a letter
within a permutation is not allowed.
268. Fundamental Principle. If a certain thing can be
done in m different ways and if, when it has been done, a cer
tain other thing can be done in p different ways, then both
420
XVII, §269] PERMUTATIONS AND COMBINATIONS 421
things can be done in the order stated in m x p different
ways. For, corresponding to the first way of doing the first
thing, there are p different ways of doing the second thing ;
corresponding to the second way of doing the first thing there
are p different ways of doing the second thing ; and so on for
each of the m different ways of doing the first thing. There
fore there are m x p different ways of doing both things in
the order stated. This fundamental principle may at once be
extended to the following form.
If one thing can be done in m ivays, and if, when it has been
done, a second can be done in p loays, and if when that has been
done, a third can be done in q ways, and so forth, then the number
of ways ill which they can all be done, taking them in the order
stated, is m X p xq •••.
Example 1. There are five trails leading to the top of Mt. Moosilauke,
N. H. In how many ways may I go to the top, and return by a different
trail ?
There are five ways I may go to the top and for each of these there are
four ways I may descend. Therefore, the total number of ways in which
I may make the round trip is 5 x 4 or 20.
Example 2. How many even numbers of two unlike digits can be
formed with the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 ?
The digit in the units' place can be chosen in any one of 4 ways and the
one in the tens' place can then be chosen in 8 ways. Tlierefore, 4 x 8 or 32
even numbers with two unlike digits can be formed from the given digits.
269. The Number of Permutations of n Different Things
Taken r at a Time. The problem of finding the number of
permutations of n different things taken r at a time can be
stated as follows :
Find the number of ways in which we can fill r places when
we have n different things at our disposal.
The first place can be filled in n ways, for we may take any
one of the n things at our disposal. The second place can
422 MATHEMATICAL ANALYSIS [XVII, § 269
then be filled in ?i — 1 ways, and hence the first and second
places together can be filled in n {n — 1) ways. Why ? When
the first two places are filled, the third can be filled in w — 2
ways. Reasoning as before, we have that the first three places
can be filled in n(n — l)(n — 2) ways. Proceeding thus, we see
that the number of ways in which r places can be filled is
n (ii — l)(?i — 2) ..• to r factors,
and the rth factor is n — (r — l) or n — r + 1. Therefore the num
ber of permutations of n different things taken r at a time is
(1) ^p^ = „(„_i)(n2)...(nr + l).
Corollary. If r = n, we have
(2) ^Pn = n (n  1) (n  2) ... 3 . 2 . 1 = n ! *
Example. Three students enter an oflBce in which there are five
vacant chairs. In how many ways can they be seated ?
Here n = 6, r = 3. Hence sPs = 6 • 4 • 3 = 60 ways.
270. The Permutations of n Things not all Different. The
number N of permutations of n things taken all at a time, of
which p are alike, q others are alike, r others alike, and so on, is
(3) N= ~ ■'
Suppose the n things are letters and that p of them are a,
q of them h, r of them c, and so on.
Now, if in any of the N permutations we replace the p a's
by p new letters, different from each other and also from the
remaining n — p letters, then by permuting these p letters
among themselves without changing the position of any of the
other letters we can form p ! new permutations. Therefore if
this were done in each of the N permutations, we should
* The product of all the integers from 1 to n is called factorial n, and is de
noted by the symbol n ! or [n. Thus 3 ! = 1 • 2 • 3 = 6. A table of the values
of n ! up to n = 10 will be found at the end of the book.
XVII, § 270] PERMUTATIONS AND COMBINATIONS 423
obtain Npl new permutations. In the same manner, if we
replace the q 6's by q new letters differing from each other and
the remaining n — q letters, the r c's by r new letters differing
from each other and from the remaining n — r letters, and so
on, we then obtain N plqlrl "> new permutations. But the
things are now all different and may be permuted in n I
ways. Therefore J^  p\  q I  r \ "■ = nl, or
p Iqlrl '"
Example. How many different permutations of the letters of the
word Mississippi can be formed taking the letters all together ?
We have 11 letters of which 4 are s, 2 are p, 4 are i. Therefore the
number of permutations is 11 !/(4 ! 4 ! 2 !) = 34650.
EXERCISES
1. If there are six letter boxes, in how many ways can two letters be
posted if they are not both posted in the same box ? Ans. 30.
2. If there are six letter boxes, in how many ways can two letters be
posted ? Ans. 36.
3. Two dice are thrown on a table. In how many ways can they
fall? Ans. 36.
4. Two coins are tossed on a table. In how many ways can they fall ?
5. In how many ways can five coins fall on a table ?
6. How many different permutations can be formed by taking five
of the letters of the word compare ?
7. Find the number of permutations that can be made from all the
letters of the word (a) assassination; (b) institutions; (c) examination.
8. Given the digits 1, 2, 3, 4, 5, 6, 7, 8, 9. Find
(a) How many odd numbers of two digits each can be formed, repeti
tion of digits being allowed.
(b) The same as (a), except that repetition of digits is not allowed.
9. How many even numbers less than 1000 can be formed with the
digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, repetition of digits not being allowed ?
10. In how many ways can a hand of ten cards be played one card
at a time ?
11. In how many ways can 3 different algebras and 4 different geom
etries be arranged on a shelf so that the algebras are together?
424 MATHEMATICAL ANALYSIS [XVII, § 271
271. The Number of Combinations of n Different Things
Taken r at a Time. TJie number of combinations of n different
things taken r at a time is
(4) „C, = "("^)("^)("' + ^).
r !
For, each combination consists of a group of r different
things which can be arranged among themselves in r ! ways.
Therefore ^C^ • r I is equal to the number of permutations of n
different things taken r at a time ; that is, „C^ • r ! = „P,, or
^ _ n(n — l)(n — 2) >» (n — r + 1)
^~ r!
Corollary 1. The value of „(7, may be written in the form
rl{n — r)l
This follows immediately from (4) if we multiply numer
ator and denominator by (n — r) !, since
n(n l)(n  2) ... (w  r + 1) • (n  r) ! = n!.
Corollary 2. The number of combinations of n different
things taken r at a time is equal to the number of combina
tions of n different things taken (n — r) at a time.
= „a.
(n — r)l (??, — [n — r]) ! (n —r) I r\
The total number of ways in which a selection of some or all
can be made from n different things is 2" — 1. For each thing
may be disposed of in two ways, i.e. it may be taken or it may
be left. Since there are n things, they may all be disposed of in
2" ways. But among these 2" ways is included the case in
which all are rejected. Therefore the number of ways of
making the selection is 2" — 1.
XVII, § 271J PERMUTATIONS AND COMBINATIONS 425
Example 1. In how many ways can a committee of 9 be chosen from
12 people ?
The required number is
^ _ ^ _ 12 . 11 . 10 _ „„^
12t^9 — 12^3 — — q — r — — — ^IM.
Example 2. From 6 men and 5 women, how many committees of 8
each can be formed when the committee contains (1) exactly 3 women ?
(2) at least two women ?
(1) The men may be chosen in eCs ways, the women in gCs ways.
The number of ways in wliich both groups may be chosen together is
eCseCs, or60.
(2) Since each committee is to contain at least three women, it can be
made up as follows : , . r ■, ^
(a) 5 men and 3 women.
(b) 4 men and 4 women.
(c) 3 men and 5 women.
Therefore the number of possible committees is «
6^6 X 6<73 + 6^4 X 6^4 + 6C3 X 5^5 = 165.
EXERCISES
1. Find 10 Cg; 11 10; 100C99.
2. How many different committees of 6 men can be chosen from a
group of 20 men ?
3. There are 20 points in a plane, of which no three are in a straight
line. How many triangles may be formed each of which has three of
these points for its vertices ?
4. How many planes may be determined by 25 points, no four of which
are coplanar, if each of the planes is to contain three points ?
5. How many different committees, each consisting of 5 republicans
and 4 democrats, can be formed from 10 republicans and 8 democrats ?
6. From 20 men how many groups of 11 men each can be picked ? In
how many of these groups will any given one of the 11 men be ?
7. Out of 6 different consonants and 4 different vowels, how many
linear arrangements of letters each containing 4 consonants and 3 vowels
can be formed? Ans. ^Ga x 403 x 7 !.
8. From ten books, in how many ways can a selection of six be made,
(1) when a specified book is always included ?
(2) when a specified book is always excluded ?
426 MATHEMATICAL ANALYSIS [XVII, § 272
272. Probability. If an event can happen in h ways and fail
in / ways, and if each of these f+h ways is equally likely, the
(mathematical) probability * of the event happening is
h
h\f
and the probability of its failing is f/{h +/). An equivalent
way of stating that h/(Ji \f) is the probability of an event
happening is to say that the odds are h to / in favor of the
event or / to 7i against the event.
The probability of an event happening plus the probability of
its failing is always equal to unity.
Example 1. Suppose from a bag containing 3 red balls and 5 black
ones, a ball is drawn at random, then the probability of its being red is f
and of its being black f . The chance that the ball is either red or black
is I + f = 1, or certainty.
Example 2. From a bag containing 3 red balls and 6 black ones, two
balls are drawn. Find the probability that (1) both are red, (2) both are
black, (3) one is red and one is black.
Two balls can be drawn in g C^ or 28 ways. Two red balls can be drawn
in 3C2 or 3 ways. Therefore the probability of drawing two red balls is
3/28.
Two black balls can be drawn in 5 (72 or 10 ways. Therefore the prob
ability of drawing two black balls is 10/28.
The number of ways of drawing one red ball and one black one is
zCi X 6^1, or 15. Therefore the probability of drawing a red and a black
ball is 15/28.
Example 3. Find the probability of throwing six with two dice. The
total number of ways in which two dice can fall is 6 x 6 or 36. A throw
of 6 can be made as follows: 1, 6; 5,1; 4,2; 2,4; 3,3; i.e. in
5 ways. Therefore the probability is 5/36.
* The reason for the definition of mathematical probability may be made
clear from the following considerations. Suppose a coin were tossed vi times
and fell heads h times and tails/ times. If n is a finite number, h and/ will
in general not be equal. But as n is increased, h/{h+f) and f/(h\f) will
approach nearer and nearer to 1/2, and thus we take 1/2 to be the probability
of the coin falling heads
XVII, § 272] PROBABILITY 427
EXERCISES
1. In a single throw with one die, find the probability of throwing an ace.
2. In a single throw with two dice, find the probability of throwing a
total of five ; six ; seven ; eight.
3. In a single throw with two dice, find the probability of throwing at
least five ; six ; seven ; eight.
4. A bag contains 5 red balls, 6 green balls, 10 blue balls. Find the
probability that, if 6 balls are drawn, they are (a) 2 red, 2 green, 2 blue;
(b) 3 green, 3 blue; (c) 5 red, 1 green ; (d) 6 blue.
5. Four coins are tossed. Find the probability that they fall two heads
and two tails. Ans. 3/8.
6. In a throw with two dice, which sura is more likely to be thrown,
6 or 9 ?
7. Find the probability of throwing doublets in a throw with two dice.
8. Five cards are drawn from a pack of 52. Find the probability that
(«) there is one pair. [Two like denominations make a pair, for ex
ample, two aces.]
(6) Find the probability that there are three of a kind ; (c) two pairs ;
(d) three of a kind and a pair ; (e) four of a kind ; (/) five cards of one
suit.
9. Four cards are drawn from a pack of 52. Find the probability
that they are one of each suit.
10. Seven boys stand in line. Find the probability that (a) a partic
ular boy will stand at an end; (6) two particular boys will be together ;
(c) a particular boy will be in the middle.
11. A and B each throw two dice. If A throws 8, find the probability
that B will throw a higher number.
12. Find the probability of throwing two 6's and one 5 in a single
throw with three dice. *
13. In tossing three coins find the probability that at least two will be
heads.
14. If the probability that I shall win a certain event is , what are the
odds in my favor ?
15. Find the probability of throwing an ace with a single throw of two
dice. Ans. 11/36.
16. Which is more likely to happen, a throw of 4 with one die or a
throw of 8 with two dice ?
428 MATHEMATICAL ANALYSIS [XVII, § 273
273. The Binomial Theorem for Positive Integral Ex
ponents. Consider the product
(x \a)(x 4 a) ••• (x + a) [to n factors]
where n is any positive integer. One term' of the product is
oj" ; it is obtained by taking the letter x from each parenthesis.
There will be n terms x^'^a, for the letter a may be chosen from
any of the n parentheses which can be done in „0i = n ways.
There will be „(72 terms x^'^a"^, for the a's may be chosen from
two of the n parentheses and the x from the remaining n — 2
parentheses. In general, there will be nO^ terms a;""'" a**, for
the a's may be chosen from any r of the n parentheses, and the
x's from the remaining n — r parentheses. Therefore
(6) (x + a)^ =zx^ + nCiX'»ia + nCax'*^ a^ + ...
\„CrX'^rar + ... + a".
This formula for expanding (x \ ay is known as the binomial
theorem. Since „(7^ = ^(7„_„ it follows that the coefficients of
any two terms equidistant from the beginning and the end are
equal. If we write — a in place of a we have
(x  ay = x~' + nCiX' ( a) + n^aaj'*'' ( ay + ... f ( a)%
or
(aja)« = aj"— „Oia."ia+ ^OgX^^a^— ^CaOJ^Vi^ j 1 ( _ 1)" a\
Example 1. Expand (2x — yy.
= 32 x5  80 x^y + 80 xV _ 40 x'Y + 10 ay^ _ yS^
Example 2. Find the sixth term of (2 as — 3 yy.
The sixth term is gCs (2xY (  3^)5, or  108,864 x^.
Example 3. Find what term contains x" in the expansion of ( x^ — J .
Call it the «"i term. ThenioC«_i(x2)"<^iy~^ is the term. In this term
we want the exponent of x to be 11. Therefore 22 — 2«— «fl = ll, or
t = 4. The coefficient of this term is  10C3 = — 120.
XVII, § 273] BINOMIAL THEOREM 429
EXERCISES
Expand the following by the binomial theorem :
1. (xl)5. 3. {2xyy. 5. (xy,
/ l\io ^ ^'
2. (2x + y)6. 4 [^) • 6. {z  xy)K
7. (0.9)6. [Hint. 0.9 = 10.1.] 8. (0.99)3.
Write down and simplify :
9.. The 8th term of (x  iy\ 12. The 6th term of (2 a; + 3 y)i2.
10. The 5th term of (2 x  y)io. 13 The middle term of (1  xy\
U. The middle term of (2 x  y)i*.
15. The middle terms of {z  \/z) i^.
11. The 7th term
Find the coefficient of
\b ix)
16. x^ in the expansion of (x^ — j ,
17. a;i8 in the expansion of [x^l] .
1 \
15
18. x^9 in the expansion of ( x* + i .
19. x^'' in the expansion of ( x* — ^ ] .
V x^J
20. By considering the expansion of (1 + 1)», prove that
21. Prove 1 _ „Ci + „(72  „C3 + •. + ( 1)" „Cn = 0.
MISCELLANEOUS EXERCISES
1. In how many ways can 10 boys stand in a row ?
2. In how many ways can ten boys stand in a row when
(a) A given boy is always at a given end ?
(6) A given boy is always at an end ?
(c) Two given boys are always together ?
(d) Two given boys are never together ?
3. How many numbers of three digits each can be formed from the
digits 1, 2, 3, 4, 5, 6, 7, when
(a) A repetition of digits is allowed ?
(6) A repetition of digits is not allowed ?
430 MATHEMATICAL ANALYSIS [XVII, § 273
4. How many numbers of three digits each can be formed with the
digits 2, 3, 5, 6, 7, 9, when
(a) The numbers are less than 500 and a repetition of digits is
allowed ?
(&) The numbers are greater than 500 and a repetition of digits is
not allowed ?
5. In how many ways can a consonant and a vowel be chosen from
the letters of the word vowels ?
6. Find n when,
(a)„(72 = 45; (6) „P3 .= 210 ; (c)„(72 = „C3. *
7. Show that the number of ways in which n things can be arranged
around a circle is (n — 3) ! .
8. In how many ways can 6 people sit around a round table ?
9. How many signals can be made by hoisting 7 flags all at a time one
above the other, if 2 are blue, 3 are white, and the rest are green ?
10. How many different numbers of seven digits each can be formed
with the digits 1, 2, 3, 4, 3, 2, 1, the second, fourth, and sixth digits being
even?
11. How many handshakes may be exchanged among a party of 10
students if no two students shake hands with each other more than once ?
12. A lodge has 50 members of whom 6 are physicians. In how many
ways can a committee of 10 be chosen so as to contain at least 3
physicians ?
13. A crew contains eight men ; of these three can row only on the
port side and two only on the starboard side. In how many ways can the
crew be seated ?
14. Find n when „+2C4 = llnC2
15. In how many ways can 18 books be divided into two groups of 6
and 12 respectively ? Atis. isC'e.
16. In how many ways can 12 students be divided into three groups
of 4, 3, 5, respectively ?
17. How many different amounts can be weighed with 1, 2, 4, 8, and
16 gram weights ?
18. How many sums of money can be made with 5 onecent pieces,
4 dimes, 2 half dollars, and 1 fivedollar bill ?
19. In how many ways can four gentlemen and four ladies sit around
a table so that no two gentlemen are adjacent ? Ans. 144.
XVII, § 273] PERMUTATIONS AND COMBINATIONS 431
20. Prove nO, + „Or_l = n+lC,*
21. How many dominos are there in a set numbered from double
blank to double six ?
22. A railway signal has three arms and each arm can take three
different positions. How many signals can be formed ?
23. Prove n+^Cr+i = nCr+i + 2 „0, + „ai.
24. How many combinations of four letters each can be made from
the letters of the word proportion ? How many permutations ?
Ans. 53; 758.
25. Find the probability that in a whist hand a player will hold the
four aces.
26. Find the probability of drawing a face card from a pack of 52
playing cards.
27. If two tickets are drawn from a package of 15 marked 1, 2, •••, 15,
what is the probability that they will both be marked with odd numbers ?
both with even numbers ? both with numbers less than 10 ? both with
numbers more than 10 ?
28. To decide on partners in a game of tennis four players toss their
rackets. The 2 "smooths" and the 2 "roughs" are to be partners.
What are the odds against the choice being made on the first throw ?
29. Prove that the sum of the coefficients of the odd terms of a
binomial expansion equals the sum of the coefficients of the even terms.
30. If n is an even integer, prove that there is a middle term in the
expression of (x + a)" and that its coefficient is even.
31. Provethat„Ci+2„(72 + 3„03 + ••• n„a„= w(2)"i.
32. Prove „Ci  2„C3 + 3„(73 + . ( l)"i • w • „C„ = 0.
* An application of this formula is the construction of Pascal's Triangle.
(o^o by definition will be assigned the value 1.)
oCo
iCo
iCi
2C0
aC'i
. 2C2
8^0
sCi
8C2
3^3
4C0
4C1
4C2
4C3
1
2
1
3
3
1
4
6
4
404
The formula in Ex. 20 shows that any number n+iCr is equal to the number
just above it, i.e. nCr, plus the number nCri which is to the left of nCr.. Thus
for example 4C8 = 3C3 f 362. We can, by means of this formula in Ex. 20,
write down the next row. It is
1 5 10 10 5 1
The numbers in the nth row of the table are seen to be the coefficients
of the terms in the expansion of (z{ a)^ (§ 273) .
CHAPTER XVIII
COMPLEX NUMBERS
274. Definitions. We have already had occasions to refer
to the socalled imaginary numbers. A number that arises as
the result of extracting the square root or, indeed, any even root
of a negative number is called an imaginary number. Thus
V— 2 is an imaginary, number ; the roots of the quadratic
equation ^624. 3 _ q^ yiz. ± 2V— 2, are imaginary numbers.
We have hitherto avoided the use of imaginary numbers as
far as possible. It now becomes desirable to take them defi
nitely into account, to learn how to work with them, and to
gain some knowledge of their usefulness. Indeed, one of the
primary objects of this chapter is to show that imaginary
numbers have quite as concrete an interpretation as the real
numbers, an interpretation which in many cases is of great
service in the solution of concrete problems.
The letter i is used to represent the socalled imaginary
unit; it is by definition such that i^ = — 1.
Numbers of the form ibj where 6 is a real number different
from zero, are called pure imaginary numbers.
Numbers of the form a \ ib, where a and b are real numbers,
are called complex numbers.
In the complex number a f ib, a is called the real part and
ib the imaginary part. In a real number the imaginary part
is zero ; in a pure imaginary the real part is zero. A complex
number a f &« is imaginary if 6 ^ 0.
When two complex numbers differ only in the sign of the
imaginary part they are said to be conjugate. Thus 3 + 2 t
and 3 — 2 i are conjugate complex numbers.
432
XVIII, § 276] COMPLEX NUMBERS 433
275. Assumption. We assume that complex numbers obey
the laws of algebra given in § 41. By applying this assump
tion we have symbolically for the sum and difference of the
two complex numbers a + ib and c f id,
a {lb ± (c \ id) = a ± c \ i(b ± d).
That is, to add (subtract) complex numbers, add (subtract) the
real and imaginary parts separately.
276. The Geometric Interpretation of the Imaginary Unit.
We now seek a geometric interpretation of the imaginary unit
i. To this end we recall the familiar representation of the
real numbers as directed segments on a line, o j
together with the interpretation of multiplica ^^co^a a
tion by  1 (§ 35). To multiply a real number ^''' ^^^
a by — 1 is equivalent geometrically to a rotation about the
point through two right angles of the segment OA which
represents a (Fig. 237).
Now, by definition, i is such a number that i^ = — 1. To
multiply a real number a by — 1 is then equivalent to multi
plying it by P, i.e. by i • i. Multiplying a real number a by i
may, therefore, be interpreted geometrically as an operation
which when performed twice is equivalent to a rotation about
in the plane through two right angles ; i.e.
to multiply a by i may be interpreted geo
metrically as equivalent to rotating OA about
" ^ in the plane through one right angle.
The number ai will then be represented by
a segment OB equal in length to OA whose
direction makes with that of OA an angle of
90° (see Fig. 238). In the figure we' have also indicated the
result of multiplying ci by i^=i 'i=—l and by i^=i • i • i= —i.
2f
434 MATHEMATICAL ANALYSIS [XVIII, § 276
Multiplying by i* = i • i . i . i = 1*2 . /2 = 1 is then to be inter
preted as a rotation through four right angles.
EXERCISES
Give the conjugates of the following complex numbers :
1. 3 + 2 I. 2. 3  4 1. 3. _ 5  3 1. 4. _ 8 + i.
Simplify the following expressions :
6. 2(3 + 4 i)  4(1  i). 7. ll^ii _ ^J^li.
6.  4(1  i) } 6(3  28 i). 8. x + iy + ix + y.
9. Prove that the sum of two conjugate complex numbers is a real
number.
10. Is the following statement true ? If the sum of two complex
numbers is a real number, the complex numbers are conjugates. Explain.
11. Prove that every even power of i is equal to either 1 or — 1.
12. Prove that every odd power of i is equal to either i or — i.
13. Find the value of i 4 2 i^ + 3 i^ + 4 i*.
14. Find the value of i^^ + i*^ + ^es _. j69 _,. 444.
277. Vectors in the Plane. We have seen that, if any real
number a is represented by a horizontal segment directed to
the right or left according as the number a is positive or
negative, then the imaginary number ai may be represented
by a vertical segment directed upward or downward accord
ing as a is positive or negative. This suggests the possi
bility of representing other complex numbers by segments
having other directions in the plane. Such a directed seg
ment will represent a magnitude (the length of the segment)
and a direction. Therefore such a segment can be used to
represent a variety of concrete quantities that are not merely
geometric; e.g. a force of a given magnitude and acting in
a given direction; a velocity, meaning thereby the speed
(magnitude) and the direction in which a body moves ; etc.
Such quantities having both direction and magnitude are
XVIII, §278] COMPLEX NUMBERS 435
called vectors, and, if the directions are restricted to lie in
the same plane, they are called plane vectors. Any plane
vector may, then, be represented by a directed linesegment
in the plane.
Two vectors are said to be equal if and only if they have the
same magnitude and the same direction. Hence, from any
point in the plane as initial point, a vector can be drawn equal
to any given vector in the plane.
278. Addition of Vectors. The addition of vectors in the
plane proceeds according to a definition analogous to the geo
metric addition of directed linesegments discussed in § 35. If
we are given two vectors AB and BC, we may
conceive the first to represent a motion from
Ato B and the second a motion from B to O.
The sum of the two vectors then represents, ^ ^
by definition, the net result of moving from A ^^^' ^^^
to B and then from B to C, i.e. the motion from A to C. The
sum of the vectors AB and ^C is then the vector AC (Fig.
239). In symbols ^s + BO = AC.
In other words, the sum of two vectors is the vector from the
initial point of the first to the terminal point of the second,
when the vectors are so placed that the initial point of the
second coincides with the terminal point of the
A,^^^^\/ first. From this definition it follows immediately
^ that, if two vectors issue from the same point
0, their sum is the diagonal, issuing from 0, of
the parallelogram of which the two given vectors form two
adjacent sides (Fig. 240).^
* If the vectors represent two forces, this shows that the sum of the vectors
represents the resultant of the forces according to the law known as " the
parallelogram of forces."
436
MATHEMATICAL ANALYSIS [XVIII, § 279
279. The Components of a Vector. The projection of a
vector on a given line is called its component parallel to the line.
Thus in Fig. 241 the directed segment M1M2 is the horizontal
»«
r ^
'
*l
1
i 1
M, M,
Fig. 241
component of the vector AB, and the directed segment ^1:^2
is its vertical component. Moreover,
vector AB — vector N^N^ + vector M^M^.
If the horizontal and the vertical components of a vector are
known, then the vector is known. Why ?
280. The Complex Number x + iy and the Points in the
Plane. Let OP (Fig. 242) be any vector issuing from 0, and
let the horizontal vectors issuing from be represented by the
positive and negative real numbers (and zero). We have seen
F;
p
Vi
■^
X J
/ V
Fig. 242
that the numbers of the form ai can be represented by the ver
tical segments issuing from 0. Here a is a real number and i is
a vector of unit length. The horizontal component of OP will
then be a certain real number x, and the vertical component a
certain pure imaginary number iy. The vector OP will then
be equal to the sum of these two components, i.e.
OP^x\ iy.
XVIII, § 280]
COMPLEX NUMBERS
437
Conversely, every number of the form x \ iy represents a
definite vector in the plane. If its initial point is at the origin
of a system of rectangular coordinates (with equal units on
the two axes), its terminal point is the point (x, y).
We have hitherto used vectors in the plane to represent the
complex numbers. If we think of these vectors as all having
their initial points at 0, each vector determines uniquely, and
is uniquely determined by, its terminal point. Hence, we can
also use a complex number to represent a point in the plane,
viz. the number x + iy will represent the point whose rectan
gular coordinates are (cc, y).
Example 1. Represent by means of vectors the complex numbers
'2 ■+ 2i and 1 ■\Qi. Find the vector that represents their sura.
In Fig. 243 the vector OA represents the complex number 2 + 2 i, and
the vector OB represents the complex number 1+6 i. The sum of these
two complex numbers is represented by the vector OC. Why ?
p
■e
^t
r, ^!i
^^ 71
' it
_, , '
 ===f K"' =
: :::Si " 
irM t
Z^ t
1 t
it ^A
t 4
l^X
0  ^
1
_x
T
/y
>
\\
\
\
\
"^
\
h ^i
^\
q2I.
N
Y
^v
^ \
A^
''
^
B
•

Fig. 243 Fig. 244
Example 2. Find the vector that represents (1 + *") — (2 — 3 1).
To find this vector, find the vectors, OA and 0J5, that represent 1 + i, and
2 — 3i, and determine OC so that OA is the diagonal through O of the
parallelogram of which OB and OCare adjacent sides (Fig. 244). Note
that the vector OC is equal to the vector BA.
438
MATHEMATICAL ANALYSIS [XVIII, § 281
281. Equal Complex Numbers. Ifx{iy = 0, then x = and
y = 0. For, if cc + iy = 0, and y=^0, we should have x/y = — i,
which is impossible. Why ?
If Xy + lyi = X2 + iy2, then x^ = x^ and y^ = 2/2 ^oi*) ^J trans
posing terms, we have (x^ — Xo) + i(yi — 2/2)= ^ Hence, we
have Xi = X2 and y^ = 2/2.
Thus, fwo complex numbers are equal if and only if the real part
of the first is equal to the real part of the second, and the imaginary
part of the first is equal to the imaginary part of the second.
Geometrically, two complex numbers are equal if and only if
they represent the same point.
282. The Polar Form of a Complex Number. Connect the
point P{x, 2/) (Fig. 245), which represents the complex number
X 4 iy, to the origin 0. If we let {p, 6) (p > 0) be the polar
coordinates of P(0 being the origin and OX the initial line),
then for any position of the point F we have
^ 'x = p cos 6,
,y = p sin 0.
Therefore, the complex number x \iy may
be written in the form
(2) x + iy = p(cose + zsine). (p^O.)
This form of complex number x + iy is called the polar form.
The angle 6 is called the angle or the argument, and the length
p is called the absolute value * of the complex number.
Example. Find the angle, the absolute value, and
the polar form of the complex number 2 } 1 2\/3.
Plot the complex number (Fig. 246). Now we have
p — y/x^ + y^. Hence p = \/4 + 12 = 4. Moreover
tan^=\/3, i.e. 6=00°. That is, the absolute
value is 4 and the angle is 60°. Therefore the polar
form is 4 (cos 60° + i sin 60°) . ^« 2*6
* Also sometimes called modulus.
V
(1)
Fig. 245
Y
J
Pi2+i2/3)
/
2/3
A'°
2
X
XVIII, §282] COMPLEX NUMBERS 439
EXERCISES
In the following exercises represent by vectors the numbers in paren
theses, and their sum or difference as the case may be :
1. (3 + + (4 + 20. 4 (540(2i).
2. (1 +3i)(560. 5. (3 + 2z)+(3+20.
3. 7_(5 4.3 4). 6. (44i)6.
Represent by a point each of the following complex numbers ;
7. 3 + 5 i. 9. 6 + i. 11.3+6 i.
8. 33i. 10. 53i. 12. 7 + iV2.
In the following exercises, represent by points the numbers in paren
theses, and their sum or difference as the case may be :
13. (3hO + (4 + 2 0. 16. (54 0(20
14. (lf3 0(5 6i). 17. (3 + 2 + (3t2i).
15. 7 (5 + 3?). 18. (3 + 3 05.
Find real values of x and y satisfying the equations :
19. 2x— iy = 4:y — Q — 4i. 22. ixy + ic+?/ = 5 + 4i.
20. X + ioy = y + 6 + 36 i. 23. a:2 + ?/2= 25  (3 x+4 y2b) i.
21. {Sx + Qy+2)i — Syx=8. 24. ix { iy = 4: i \ 5 x.
Find the angle and the absolute value of each of the following complex
numbers. Represent the numbers in polar form :
25. l + iV3. 27. 11. 29. 3i. 31. 8i.
26. 5 + 5i. 28 1*^. 30. 8. 32. 12 + 5 i.
2 2
33. Can the complex number x + iy, where x and y are real numbers,
equal 7 ?
34. Under what circumstances is the sum of two complex numbers a
real number ?
Change the following complex numbers from the polar form to the
form x}iy :
35. 3(cos 30° + i sin 30°) . 38. 2 V2(cos 225° + i sin 225°) .
36. 4(cos 135° + z sin 135°). 39. 4(cos 90° + i sin 90°).
37. cos 210° + i sin 210°. 40. 8(cos 180° + i sin 180°).
440
MATHEMATICAL ANALYSIS [XVIII, § 283
283. Multiplication of Complex Numbers. Our assump
tion in § 275 allows us to multiply two complex numbers
iCi f iz/i and X2 + iy^ as follows :
(a?! + *2/i)(aJ2 + m) = 3^1352 4 iy 1X2 + ixiy2 + ^^l2/2
= (aJia^2  2/1^2) + *(^i2/2 + X2yi)'
If the two numbers are written in polar form, the multipli
cation may be performed as follows :
^'1 + *2/i = pi(cos 61 4 i sin ^1),
^2 + %2 = p2(cos $2 + I sin ^2)
By actual multiplication, we have
(•^1 + iyi)ix2 + m)
= P1P2 [cos 61 cos ^2 h '' (sin Oi cos ^2 + cos Oi sin ^2) — sin Oi sin ^2]
= P1P2 [cos (^1 4 ^2) + i sin ((9i + ^2)].*
Therefore, the absolute value of the product of
two complex numbers is equal to the product
of their absolute values, and the angle of the
product is equal to the sum of their angles.
In Fig. 247 the points Pi and P2 represent the
complex numbers xi + iyi and X2 + iy2 respectively.
The point P3 represents {xi\iyi)(x2 + iyo)
Fig. 247
284. Division of Complex Numbers. The quotient of two
complex numbers Xi \ iyi and X2 f iy^ luay be reduced to the
form a + ib if we make the denominator real by multiplying
both numerator and denominator by the conjugate of the
denominator. Thus,
Xi + iyi _ a?i + m . X2 — iy2 ^ X1X2 h iyiXj — ixiy2 — i^yiyz
X2 f iyo «2 + m ^2 — iyi
^2' + 2/2'
* See § i:ts
X2^ + 2/2'
^%Vi
x^^y^
XVIII, § 284]
COMPLEX NUMBERS
441
If we write the two complex numbers in polar form and
then perform the division, we have
pi (cos $1 + I sin ^i) _ pi (cos Oi + i sin ^])(cos 0^ — ^ sin O^)
P2 (cos $2 + i sin O2) p2 (cos O2 + i sin 62) (cos ^2 — i sin ^2)
^ Pi [cos (^1  O2) + i sin (^1  ^2)1
P2(cos2^2 + sin292)
= ^^ [cos (^i  $2) + 1 sin (0,  ^2)].
Therefore, the absolute value of the quotient of two complex
numbers is equal to the quotient of their absolute values, and the
angle of the quotient is equal to the difference of their angles.
Example 1. Find analytically and graphically the product (1 f i)
Solution. Analytically,
(1 + 0(3 4. V8 i) = 3 + 3 I + V3 ^ + \/3 i^ = (3  V3) + i (3 + V3).
Graphically, writing the complex numbers in polar form, we have
V2(cos 45° + i sin 45°) and 2 V3(cos 30° + 2 sin 30°).
Therefore pi = \/2, P2 = 2 V3, Ox = 45°, 62 = 30°.
Ir 4^
t^
£
1
t
^rit 
7 p^^^'
tc^
^_
X
1 . .
Fig. 248
Hence the absolute value of the product is p\p2 =2 V6 and the angle of
product is 75°. In Fig. 248 the points Pi, Pi, and P represent respectively
the complex numbers 1 + i, 3 + iVS, (1 + 1)(3 + iV3).
442
MATHEMATICAL ANALYSIS [XVIII, § 284
Example 2. Find analytically and graphically the quotient
(3 + iV3)/(l + 0
Solution : Analytically :
3 + iv/3 ^ 3 + iV3 1i^ (3+ V3)i(3V3) ^
l+i
1 + t 1  i
z
^x
c^ t
//^
gr ^_
jo^^ ^±:
■E
Fig. 249
Graphically^ using the results in Ex. 1, we see in Fig. 249 that the
points Pi, P2, and P represent respectively the complex numbers
(1 + 0, (3 + iV3), (3 + iV3)/(l + 0.
EXERCISES
Perform the following operations analytically and graphically :
1. (1 + 0(2 + 2 0. _ g lzV3
2. (1 + i V3)(2 4i2V3). * 3 '
3. (2 0(5) T 5 + 5t
4. (i_j0(220( 1 + iV3). * l»
3 + iV3 8. n0' 
6.
1 + I 2 + I 2 \/3
Perform the following operations analytically :
a 3 + i U. (i9 + iio + i" + ii2)7.
l_^18t 329t
10.
7iV2
3 + 4 i 3  4i
11.
12.
13.
(2 + 1)2 (2  0'
X + t Vl — X2
16.
V V2 /
Vlx^
17 2 + 3 ;; 3 + 2 ^
3 _ 4 I 3 + 4 i
1 + i
18. V? + 24 i.
XVIII, § 285] COMPLEX NUMBERS 443
285. DeMoivre*s Theorem. The result of § 283 when ap
plied to the product of any number of complex numbers leads
to the following :
I. The absolute value of the product of any number of complex
numbers is equal to the product of their absolute values.
II. Tlie angle of the product of any number of complex numbers
is equal to the sum of their angles.
If the above statements be applied to a positive integral
power of a number, i.e. to the product of n equal factors,
we obtain
(3) [/3(cos 6 \ i sin ^)]" = p"(cos nd \ i sin n 6).
For the special case p = 1 we obtain
(4) (cos 6 + i sin 6)" = cos n 9 + / sin n 6.
This relation we have just proved for the case where n is a
positive integer. It also holds when ti is a negative integer.
For we have
(cos^ + isin^r= i ^cos^tsin^
^ cos + 2 sin e cos2 6 f sin^ 6
= cos (— 0) 4 i sin (— 0),
and hence
(cos + i sin 6)'' = [cos (— 0) f i sin (— 0)^
= cos (— p 6) 4 i sin ( — p 0).
Further, \i n — 1/g, where g is a positive or negative integer,
we have, by what precedes,
(5) (cos + i sin Of = [Uo^ + i sinVl^
and hence
(6) (cos ^ + i sin Oy = (^cos  f / sin ^\ = cos ^ ^ + / sin ^ 9.
6
13= COS+ isin,
q q
444 MATHEMATICAL ANALYSIS [XVIII, § 285
This shows that relation (4) is valid for all rational values ofn.
It should be noted, of course, that relation (5) states merely
that a certain q^^ root of cos 6 + i sin 6 is cos {O/q) f i sin {6/q)
and that a similar statement applies to relation (6). The fact
expressed by (4) is known as De Moivre's theorem.*
286. "Powers and Roots of Numbers. De Moivre's theo
rem often enables us to compute an integral power of a complex
number without difficulty, as the following example will show.
Example 1. Find the value of (2{2i)^. The polar form of this
number is 2 \/2(cos 46'^ + i sin 45") . Hence
(2 + 2 0^= (2\/2)6(cos225° + i sin 225°)
= 128V2( 7:^= 128 128 I.
V V2 V2y
To find the nth roots of a number requires special methods.
Example 2. Find the 5th roots of 2 + 2 i.
Here as in Ex. 1 we may write
2 + 2 i = 2V2(cos 45° + i sin 45°)
and hence (2 +2 0^=(2V2)^(cos9° + isin9°).
But this is not the only number whose fifth power is 2 +2 i. For we
may write 2 + 2i = 2v'2[cos(45° + k 360°)+ i sin(45° + fc 360°)], where
k is any integer. That is to say,
(2 + 2i)^=(2\/2)^[cos(9°+ ^•72°)+ isin(9° + Ar72°)].
For the values A; = 0, 1, 2, 3, 4 we get the five numbers
(7)
(2\/2)^"(cos9° + isin9°), (2 V2)^ (cos 225° + i sin 225°),
(2\/2)^^(cos81° + i sin 81°), (2V2)^(cos297° + i sin 297°).
[ (2 V2)^(cos 153° + i sin 153°),
The succeeding values of k (i.e. A = 5, 6, •••) evidently give numbers
equal to the preceding respectively. Each of the five numbers is a fifth
root of 2 + 2 i ; they are all different.
♦Abraham de Moivre (lfi671754), a mathematician of French descent
who lived most of his life in England.
XVIII, § 286] COMPLEX NUMBERS 445
The general formulation of the problem of finding the nth
root of a number z = /o(cos 6 \i sin 9) is as follows. The most
general form for z is
z = p[Gos{e + k 360°) 4 i sin((9 + k 360°)],
where k is an integer.
This gives, by De Moivre's theorem,
1 1
z'==p'
(9 + A: 360° , . . 6> + A; 360
cos ^ h t sm —^
The n values A: = 0, 1, 2, —, n — 1 give n different values for
2^/" and no more values are possible. Why? Here pV« means
the numerical nth. root of the positive number p. We have
then: Every com2i>lex number (^0) has just n nth roots. These
n roots all have the same absolute value ; their angles may be
arranged in order in such a way that every two successive
ones differ by 360°/7i.
EXERCISES
By using De Moivre's theorem find the indicated powers, roots, and
products.
1. (4 4i4V2)6. 4. (3 + iV3)io.
2. (cos 10° + i sin 10°)9. 5. (  1  WS)^.
3. (l+iV^. 6. (2+2«)*.
7. [3(cosl5° + isinl5°)]i5.
8. [2(cos 20° + I sin 20°)][3(cos 70° + i sin 70°)].
9. [2 42i][V8 + t].
10. (33 0(l + iV3).
11. \/4 + i4>/2.
12 V3 + tV3.
13. </_4+4z.
15.
V liV3.
16.
v^cos 45° + i sin 45^.
17.
V2Ti.
18.
The cube roots of 1.
19.
V:^.
20.
V2^.
14. \/8(cos6U° + isin60^).
21. Prove that tlie n nth roots of a given number z are represented by
the vertices of a regular polygon of n sides whose center is at the origin.
446 MATHEMATICAL ANALYSIS [XVIII, § 287
287. Applications in Trigonometry. De Moivre's theorem
may be used to advantage in certain trigonometric problems.
I. To express cos nO and sin nd in terms of cos and sin 6.
' We have the relation
cos nO + i sin nO = (cos $ { i sin Oy
= cos" Ohn i cos"i ^ sin ^ f ^K^^ — ^) ^2 cos"2 sin^ ^ + ...
A
If in this relation we equate the real and the imaginary
parts we get the expressions desired.
Example 1. Express cos 6 d and sin 6 ^ in terms of cos 6 and sin d.
The above method yields in this case :
cos 6 ^ + i sin 6 ^ := (cos 6 ■{■ i sin 0)'^
= cos6 e \Q icos^ ^sin ^  15 cos* 0sin2 ^  20 1 cos^ 6 sin' 6 + 16cos2 ^sin* d
+ 6 I cos 9 sin° 6 — sin^ d.
Equating the real parts, we have
cos 6 £> = coss e — 15 cos* e sin2^ + 15 cos'^ sin* d — sin^ 9.
Equating the imaginary parts we get (after dividing by i)
sin 6 ^ = 6 cos5 d^inO — 20 cos^ d sin^ ^ + 6 cos ^ sin^ d.
II. To express cos"" 6 and sin"" 6 in terms of sines and cosines
of multiples of 0. If we place w = cos + i sin 0, we have
u^ = cos kO 4 i sin kO, u~^ = cos kO — i sin kO.
Adding and subtracting these equations, we have
.r,. [ w* + w'' = 2 cos k$,
[ u^ — n * = 2 4 sin kd,
for any integral value of k.
In particular when /c = 1, we have
2 cos 6 = u ^ u~^, 2 i sin $ = u — ^^~^
It follows that
2^ cos" e= {u ifci)«= w"4 nu'^{ ^^^'~^\ ''*\ \na^''^^+u'\
XVIII, § 287] COMPLEX NUMBERS 447
The fact that the coefficients in the binomial expansion are
equal in pairs makes it always possible to group the terms as
2'^ cos"^ = {w H ?r") h n(ifc'^2 f ir(«2))+ ....
But the terms in parentheses on the right are equal respec
tively to 2 cos ?i^, 2 cos {n — 2)0, •••. The following examples
will make the method clear.
Example 2. Express cos* in terms of cosines of multiples of 6.
We set
= w4 + 4m2_.6 + 4 u^ + M*
= (m* + M*) + 4(m2 __ u2) f 6
= 2 cos 4 ^ + 4 . 2 cos 2 + 6.
Dividing both members by 2* we obtain the desired result
cos* e = I (cos 4 ^ + 4 cos 2 ^ + 3) .
Example 3. Express sin^ 6 in terms of multiples of the angle d.
We set
25 1*5 sin^ e = {u— u^y
or
32 i sin^ 6 = u^ — bu^ \10 u— Id ir^ + 5 m^ — u^
= («5 _ <5) _ 5(1(3 _ «3) 4. 10(w _ M1)
= 2 I sin 5 ^  5 • 2 i sin 3 ^ 4 10 . 2 i sin d.
Whence
sinS d = ^^ (sin 5 ^ — 5 sin 3 ^ + 10 sin ^) .
EXERCISES
Express each of the following in terms of cos d and sin 0.
1. cos 2 and sin 2 ^. 3. cos 4 ^ and sin 4 ^.
2. cos 3 e and sin 3 ^. 4. cos 5 ^ and sin 5 0.
6. Show that tan 4 g ^ ^ <^^^ ^ (^  ^^"' ^) .
16 tan2 + tan* ^
6. Find tan 5 ^ in terms of tan 0.
Express each of the following in terms of sines and cosines of multi
ples of ^ :
7. sin8 0. 9. sin* 0. 11. cos^ 0.
8. cos8^. 10. cos«^. 12. sin6^.
418 MATHEMATICAL ANALYSIS [XVIII, § 287
MISCELLANEOUS EXERCISES
Solve the following equations and illustrate the results graphically.
1. x3  1 = 0. 3. x5  32 = 0. 5. cc8  1 = 0.
2. x3 + 1 = 0. 4. a;6  1 = 0. 6. a:^ + 1 = 0.
7. Prove that
cos nd = I [cos d + I sin ^]" + ^[cos 6 — i sin ^J".
8. Prove that
i sin nd = I [cos 6 \ i sin ^]" — i [cos — i sin ^J**.
9. Prove that
/I f sin ^ + t cos ^X'* ,, /,N , • • /T „N
{ '— — =cos(lmr—nd)+ismUn'ir — nd).
10. Prove that the product of the n nth roots of 1 is 1, if n is odd, and
— 1 if n is even.
11. Prove that the sum of the n nth roots of any number is 0.
12. Complete the discussions in § 287 to derive the following formulas.
L (a) cos;i^=cos"0— ^^^^i^~!)cos"'^^sin2^
^ ^ 2!
, n(n— l)(n — 2)(n — 3) „ a ^ ■ a ^ .
_l — V ZA zs 1 cos "4 d sm^ 6 \ •••.
4 1
(6) sin nd=:n cos'»i ^ sin ^  ^(^  ^) ( ^.ZL^ cos^s q gin^ ^
o !
, nCn l)(n  2)(n3)(n4) „ ,^ . ,^
+ ~^^ ^^ — — ^ cos"fi e sin^ ^4 . . •.
5 !
II. (a) cos«^=l rcosn^+ncos(n2)^+*^— ^cos(n4)^+"l.
n
(b) sin»^ =^=^1 cosn^ ncos(n  2)^ +^^^=^cos (n4)^+ •••],
if n is even ; but
sin"/? = i=J) ^ rsinn^nsinrn2)^4 ^^^~^^ sii
if n is odd.
fsin «^n sin(n2)^ + ^^^ ^^ sin (n  4)^+ ...],
CHAPTER XIX
THE GENERAL POLYNOMIAL FUNCTION
THE THEORY OF EQUATIONS
288. The General Polynomial Function of Degree n. The
general polynomial of degree n,
f{x) = a^x"" 4 a^ix""^ + a,^2X^^ H \ aiX { a^ (a„ :^ 0),
has already been defined (§ 255). We have already dis
cussed in some detail special cases when the degree of f(x) is
1, 2, 3, (Chapters III, IV, V). For these cases we proved that
the function is always continuous, and we learned how to find
the slope of the graph of the function at any point. It is our
present purpose to extend these results and methods to a func
tion represented by a polynomial of any degree.
ax^ + a^iX""^ H h aiK + otr
289. The Slope of the Graph of /(x).
the slope of the graph of the equation
(1) y=n^)
at any point Pi{xi, y^ of this graph,
we first find the slope A^z/Aic of the
secant P^Q (Fig. 250) joining the
point Pi to any other point Q(flJi+ Aa?,
2/i + Ay) on the graph. To this end
we must first calculate the value
of Ay in terms of Xi and Aoj. We
have
2o 449
Continuity. To find
Fig. 250
450 MATHEMATICAL ANALYSIS [XIX, § 289
= a,(a^i + ^xy + a«_i(a:i + Axf'^ +  +ai(xi + Ax)\aQ.
Vi =/W= «„a;i" + a„_iaji«i + .•• + a.x^ + a^.
By subtraction and proper grouping of terms we find
(2) Ay=f(x,{.Ax)f(x{)
= alix, + Axy a^r] + a„_i[(a;i + Axy^x^^^]
H h«i[(^i + Aa;) — ajj.
Each of the terms of this expression is of the form
(3) al(x, + Axyx,'^l
and the whole expression is equal to the sum of all terms ob
tained from (3) by letting k take on the values k = n,n^l, —, 1.
Expanding the first term in the brackets, we obtain
a,l{x^ + Axy  X,'']
= «,[.T/+A;a;i*iAa;+ ^fc^a;i*2Aa;2 + ... + Aa;* a^i*]
A
^alkx^^^^~^x^''b.x +  + Aa;*i]Aa?.
Z
It is clear from this expression that for every value of k
the expression (3) has A« as a factor. Moreover the expres
sion (2) for A?/ is the sum of such terms as (3) for different
values of k ; and, since each of these terms has the factor Ax',
their sum has the factor Aa;. Hence, if we divide Ay by Aa;,
we have for the slope Ay/Aa; of PiQ, the expression
Aa/ u
f terms with higher powers of Aa;] k=.n.
+ a,_j[(n  l)a;i"2 + (^^  l)(n  2) ^^,^_3^^
4 terms with higher powers of Aa?] k = n — \.
f a2[2 a^i + Aa;] A: = 2.
h tti A: = L
XIX, § 290] GENERAL POLYNOMIAL FUNCTION 451
The slope m of the graph is the limit approached by Ay/Aa;
as Aic approaches zero (i.e. as Q approaches Pi along the curve).
This gives finally
(4) m = ?ia„iCi'*^ +(n — l)ani^i''~^ H f ^ ^2^n + «i
[Note that for the values w = 3 and w = 2 this reduces to the expres
sions previously derived for the cubic and quadratic functions. ]
Moreover, it follovt^s from the remark above, concerning the
fact that Aa; is a factor of A?/, that as Ax approaches zero, Ay
approaches zero also. But this proves that f{x) is continuous
for every value of x. We have then the theorem :
Every polynomial f{x) is a continuous function of x.
290. The Derived Function. In previous cases where we
have considered the slope of a curve y =f(x) we have always
considered its value at some given point Pi on the curve. As
the point Pj moves along the curve, however, the value of the
slope in general changes. In other words, the slope itself may
be considered as a function of a;. This function is called the
derived function or the derivative of f(;x). If the original
function is denoted by f{x), the derived function is denoted
hj f'{x). In case of the polynomial f{x) considered in the last
article the derived function f\x) is obtained from the expres
sion for the slope m by letting the given value Xi become the
variable x, i.e. if f(x) = a^x'' + cini^'"'^ H h ^i^^ + «05 ^® ^^^®
the derived function
(5) f\x)= na,x^~^+ (n  l)a„_ia;»2+ ... + a^.
The derived function of any polynomial is readily written
down from the following consideration. The derivative of any
term a^x'' is Zca^ic*"^ ; i.e, it is obtained by multiplying the term
by the exponent ofx and reducing the exponent of x byl. Thus
the derivative of a^ is 3 x^, of 10 a;^ is 20 x. The above expres
sion for/' (a;) shows that the derivative of a polynomial is the
452 MATHEMATICAL ANALYSIS [XIX, § 290
sum of the derivatives of its terms. Thus the derivative of
6 a;^ — 3 ic* + 7 a;2 — 1 may be written down at once ; it is equal
to 35 a^ — 12 a;3 _j_ 14 ^.^ Observe that the derivative of a con
stant is 0.
The relation between the derived function f{x) and the slope
of the graph at any point, is expressed as follows :
The slope of the graph of the curve y =f{x) at the point x = Xi
is equal to the value of the derived function for x=Xi^ i.e. m=f'{x^.
Further, since the derived function of a polynomial is a
polynomial, it follows from the theorem at the end of the last
article, that the derived function of a polynomial f{x) is a con
tinuous function of x.
EXERCISES]
Find/Cx) when
1. /(a;) = x3 + 4 x2  6 a; + 3.
2. /(a;) = 5a:64a;8 + 6a;2 + 2a; + l.
8. /(x) = 7 a:7  4x3 + 2 X + 19.
4. /(x)=3x64x* + 2x8 + 3x2+ 1.
6. Find the equation of the tangent to 2/ = 4 x* ~ 3 x + 1 at (1, 2).
6. Find the equation of the tangent to y = x^ — 5x2 + 2 at the point
(1,2).
291. The Graph of a Polynomial S{x). In drawing the
graph of a given polynomial of degree greater than 3, we may
proceed as in the cases of polynomials of degrees 2 and 3.
There are two general theorems to aid us :
(1) The graph of any polynomial is a continuous curve ; in
particular, the value of y does not become infinite except when
X becomes infinite.
(2) The tangent to the graph at any point P turns continu
ously as P moves along the curve ; i.e. the curve has no sharp
corners and the tangent is nowhere vertical. (Why ?)
We found in discussing the graphs of cubic functions that
XIX, § 291] GENERAL POLYNOMIAL FUNCTION 453
the values of x for wMcli the slope is zero were particularly
helpful, ill view of the fact that they gave us, in general, the
turning points (maxima and minima) of the graph. Let us
apply these principles to an example.
Example. Draw the graph ofy=f(x)= (3 a^— 4 x^  12 «2 + 3).
Weliave f(x) = i(x^ x^2x)= 4x(a; 2)(x + 1).
Hence/(x) = when a; = 0, 2, — 1.
We require next a table of correspond
ing values of x and y. Here synthetic
division is often convenient. Thus, to find
/(ic) when a; = 2, we write
3 4 12 3[2
6 41632
3 28161
29 = 32/.
Hence y = — 9 when x :
= 2.
When a; = 3, we have
3  4  12
9 15 9
33
27
3 6 3 9 30 = 32/.
Hence y = 10 when a; = 3.
:::::i^::"t:
22
^__.5.
==========;==
::::3::^:::i=:::
;::::i:::i::i:
^^^
Fig. 251
We may note that since all the partial
results 3, 5, 3, 9, 30 are positive, any value of a; > 3 will give values of y
greater than 10.
Finding the values of y for other values of x, we have the following
table:
x =
2
1
1
2
3
y =
111
1
1
3^
H
10
w =
We have also indicated in the table the values of x for which m is zero.
These data give us the graph exhibited in Fig, 251. This example sug
gests certain other general theorems regarding the graph of a polynomial,
which are discussed in the following articles.
454 MATHEMATICAL ANALYSIS PCIX, § 292
292. The Value of a Polynomial for Numerically Large
Values of x. In the example of the last article we saw that
for all values of x>S, the values oif(x) were greater than 10 ;
in fact, the nature of the synthetic division showed that as x
increased indefinitely from x = 3, the value of f{x) increased
indefinitely. Any polynomial f(x) of degree n with real coefii
cients (§ 288) may be written in the form
f{x) = aAl +f^^i=l^' + ^^=?^' +  +^^1
[_ \ ttna?" anX"" anX'^Jj
=a„x.[l+(o„_,l+c„.,l+...+c.i)].
Since the absolute value of a sum is equal to or less than
the sum of the absolute values of its terms (§ 35), we have,
X'^l
11.
<
1
Gn1
X
X'^l
1
Cn —
i(c,.i + c„_2 +  + eo)<^, (kl>l).
x\ \x\
where c is a positive number independent of x. Hence, if  a;  >c,
the value of the expression in square brackets above is cer
tainly positive. Therefore for sufficiently large values a;,
the sign oif{x) is the same as the sign of a„ic'».
If an is positive and x becomes positively infinite, f(x) is
positive. If a„ is positive and x becomes negatively infinite,
f{x) is positive if n is even, and negative if n is odd. If a^
is negative and x becomes positively infinite, f(x) is negative.
If a„ is negative and x becomes negatively infinite, f(x) is
negative if n is even, and positive if n is odd.
As X increases indefinitely in absolute value, the value of f(x)
increases indefinitely in absolute value. For sufficiently large
values of\x\, the sign off{x)is the same as the sign, of anX"".
XIX, § 293] GENERAL POLYNOMIAL FUNCTION 455
In particular, this leads us to the following theorems.
Iff{x) is a polynomial of even degree, the infinite branches of
the graph ofy = f{x) are either both above the xaxis or both below
the Xaxis (according as a„ is positive or negative).
If f(x) is a polynomial of odd degree, the infinite branches of
the graph of y =^ f(x) are on opposite sides of the xaxis (below
the a;axis on the left and above the a>axis on the right, if a,^>0 ;
above the icaxis on the left and below on the right, if a^<0).
From these theorems and from the continuity of the function
f(x) we derive the following corollary.
The graph of a polynomial f{x) of odd degree with real coeffi
cients must cross the xaxis at least once and, if it crosses more
than once, it must cross it an odd number of times. The graph
of a polynomial of even degree with ieal coefficients either does
not cross the xaxis at all or it crosses it an even rmmber of times.
293. The Zeros of a Polynomial /(x). The Roots of the
Equation /(x) = 0. A value of x for which f(x) — is called
a zero of f{x) ; i.e. if f(b)= 0, then 6 is a zero of f(x). The
zeros of f(x) are, therefore, the values of x which satisfy the
equation/ (a;) = 0. The zeros of /(a;) are called the roots of the
equation f(x) = 0. The factor theorem (§ 261) tells us that if
a is a zero of f{x), then a; — a is a factor of f(x). Since a
polynomial of degree n cannot have more than n distinct fac
tors of degree one, we may state the following theorem.
A polynomial f(x) of degree n cannot have more thann dis
tinct zeros.
Since at the turning points of f(x) the slope is always
zero, it follows from the fact that the derived function is of
degree n — 1 that a polynomial f{x) of degree n cannot have
Tnore than n — 1 turning points {maxima andminimxi).
456 MATHEMATICAL ANALYSIS [XIX, § 294
294, The Number of Roots of /(x) = 0. We have seen that
every quadratic equation has two roots which may be real or
imaginary and which may be equal. We have also seen that
every cubic equation f(x) = 0, whose coefficients are real, has
at least one real root. If this root be ri, we may write (§ 261),
f{x) = (x — rj Qix), where Q(x) is a polynomial of degree 2.
The latter has two zeros, real or imaginary, so that any cubic
function with real coefficients may be resolved into 3 linear
factors, ^(^) ^ ^^(^ _ ^^)(^ _ ,,^)(^ _ ^^y
It may be proved that any polynomial {no matter whether the
coefficients are real or imaginary) has at least one zero (real or
imaginary). This statement is called the fundamental theorem
of algebra. We shall accept it as valid without proof, since
its proof is too difficult for an elementary course.* From this
theorem it is easy to prove the following theorem :
Any polynomial f(x) of degree n may be resolved into n linear
factors.
Proof : By the fundamental theorem, f(x) has one zero.
Denote it by r^. The factor theorem then gives
f{x)^{xr,)Q,{x),
where Qi is a polynomial of degree n — 1. By the funda
mental theorem, Qi{x) has a zero r2. Hence
q,ix) = (oj  ro)Q2(a;), or f{x) = (a;  r;){x  r.,)Q2{x).
Again, Q2(x) is a polynomial of degree n — 2. If n > 2, Qg
has a zero, say r^, which leads to the expression
f{x) = {x ri) (x  rg) {x  rg) Qz{x),
where Qi(x) is a polynomial of degree n — S. Continuing this
* This theorem was first proved by Gauss in 1797 (published 1799) when he
was 18 years old. For proof see Fine, College Alyelfra, p. 588.
XIX, § 294] GENERAL POLYNOMIAL FUNCTION 457
process we find
/(a;) = (a;  ri)(a; ^ rj) "• (aJ  r„)Q„,
where Qn is a constant which evidently must be a„ if f(x) is
a^Xn + ••• + Oq. We have then finally
f(x) = a^(x  ri)(x  r.2)  (x  r„).
Each of the numbers Vi, r2, •••, r„ is a root of the equation
f(x) = 0. This proves the theorem just stated.
Moreover, no number different from Vi, rg, •••, r„ can be a root
of this equation. For suppose s were such a number, then we
should have /(s) = a„(s — ri){s — r2) ••• (s — r„). Since each
of these factors is under the hypothesis different from zero,
the product /(s) is different from zero. Some of the num
bers ri, r2, •••, r„ may be equal, however. This possibility
leads to the following definitions. If f{x) is exactly divisi
ble by a; — r but not by (a; — r)^, then r is called a simple
root of f(x) = 0. If f(x) is exactly divisible by (x — ry but
not by (a; — r)^ then r is called a dovble root of /(x) — 0. If
f{x) is exactly divisible by (a; — r)* but not by (x — r)^^ then
r is called a A:/oM root, or a root 0/ order k. A root of order
greater than one is called a multiple root. If /(a;) represents
a polynomial, the equation f(x) = is called an algebraic
equation. Then we may state the last theorem as follows :
Every algebraic equation of degree n has n roots and no more,
if each root of order k is counted as k roots.*
EXERCISES
1. Is 1 a zero of the polynomial x'' — Sx^ + 2z* — x\S?
2. Is 2 a zero of the polynomial a;* — 16 ?
3. Is 3 a root of the equation x^ + Zx'^{xBz=0?
* It is logically necessary to note the fact that, if exactly k of the roots
^i> ^2» — equal r, f{x) is divisible by (k — r)* but not by (a; — r)*+i. Why?
458 MATHEMATICAL ANALYSIS [XIX, § 294
4. Find k so that a; =1 is a root of the equation a;^ + A^x^ — x f 1 = 0.
6. Pind k so that 2 is a root of the equation x^ + x^ — Ax + 3 = 0.
6. How many roots has the equation x'^ + x8 + x + 3 = 0? How
many of these roots are positive ?
7. How many roots has the equation x^— 2 x* + x^ — 3 x2+2 x— 1=0 ?
How many of these roots are negative ?
8. Find graphically the real zeros of the functions
(a) xs  X. (6) x3 + 2 X  1. (c) x^ + 3 x + 2. (d) x^  x2  6 x + 8.
Draw the graph of each of the following functions :
9. t/ = t^3j [3 x*  4 x8  24x2 + 48 X + 13].
10. y = i^j [3 x* + 8x8  6 x2  24 X  12].
11. ?/ = ^Jj [3 X* + 4 x8  12 x2 + 24].
12. 2/ = 2 x*  14 x3 + 29 x2  12 X + 3.
13. Prove, without assuming the fundamental theorem of algebra,
that every algebraic equation of odd degree with real coefficients has at
least one real root.
295. Successive Derivatives. The derived function of a
polynomial f(x) of degree n is a polynomial f'{x) of degree
n — 1. The derivative of f{x) is a polynomial of degree
71 — 2, is denoted hy f"(x), and is called the second derivative
of f{x). Similarly, the derivative of f"{x) is called the third
derivative of f(x) and is denoted by f"'{x). Similarly, the
fourth, fifth, etc. derivatives may be found. The nth derivative
of a polynomial of degree n is evidently a constant.
Thus, if /(x)=x*3x37x + 2, we have /'(x) =4x89x2  7,
/"(x) = 12x2 18x, /'"(x) = 24x 18, /i^(x) = 24.
296. Taylor's Theorem. The following formula is known
as Taylofs theorem:
(6) /(x) = /(a) + /'(a)(x  a) +^ (x  af + ...
+m(xar.
n I
This formula enables us to express any polynomial in a? as a
polynomial in a? — a, where a is any constant.
XIX, § 296] GENERAL POLYNOMIAL FUNCTION 459
For example, if we have/(ic)= a:^ _ 4 ^ + 2 and desire to express /(x)
in terms of x + 1, we first find /'(x) = 3 x^  4 ; /"(x) = 6 x ; /'"(x) = 6.
The coefficients in the above formula are, for a = — 1,
/(l)=5, /'(1) = 1, /"(l) = 6, /'"(1) = 6.
Therefore we have, from (6),
x3 _ 3x + 4 = 5 (X + 1) 3(x + l)H(x + 1)3.
Proof. We have seen in § 290 that the derivative of a;* is
kx^~^. Likewise the derivative of {x — a)* is k(x — ay~\ For
if y z=i{x — ay, we have, as in § 289,
y + Ay ={x\ Ax— a)*=[(a;— a)+ Ace]*
= (x — a)*+ k(x — ay~'^Ax + terms with a factor Aaj^.
Hence
^ = k{x — a)^^ + terms with a factor Ax,
Ax
and the limit of Ay /Ax is obviously k(x — tt)*~^
Let us now set
(7) /(aj)=4, + A(« a) 4^2(0; a)2+  + J.,(a; a)* +.
We then have, by taking successive derivatives of both sides,
f(x) = ^li + 2 A2{x  a) f ... + kA,{x  ay' + ...,
r(x) = 2 ^2+  + A;(A^  l)A,(x  af' + ..,
/(*)(a;) = Zc ! ^;t + terms containing {x — a) as a factor.
These relations must all be true for all values of x ; hence
they must hold when x = a. But this gives
f{a)=A„ f'{a) = Au /»=2^2,, /*(«) = A: ! ^„ .
Hence
By substituting these values in (7) above, we obtain Taylor's
460 MATHEMATICAL ANALYSIS [XIX, § 296
theorem as given in relation (6). Another form of Taylor's
theorem is obtained by replacing a; by a; + a in relation (6).
This gives
(8) /(x + a) = f(a) +/'(fl)x +^^x^ + ... +i^x^.
EXERCISES
1. Write down the successive derivatives of the following polynomials :
(a) x8 + 4a;2_i2a;+l7.
(6) 2x*3a:8 + 8x214x + 18.
(c) a;5 + 2 a; 1.
(d) 1 3x +4x2 + 5x8.
2. Prove that the nth derivative of a^x" + anix"i + ••• + aix + oo is
equal to OnTi I .
3. Expand each of the following by Taylor's theorem :
(a) x3 + 4 x2 — 12 X + 17 in terms of x  1.
(6) 2 X*  3 x3 + 8 x2  14 X + 8 in terms of x  2.
(c) x^ + 2 X — 1 in terms of x + 1 .
Id) 1 — 3 X + 4 x2 + 5 x3 in terms of x + 2.
4. By relation (8) in § 296 express each of the following as a polyno
mial in X :
(a) /(xl)if/(x)=x5 + 4x212x + 17.
(h) /(x2) if/(x)=2x43x8 + 8x214x + 8.
(c) /(x + 1) if/(x)=x6 + 2xl.
W /(x + 2)if/(x) = l3x + 4x2 + 6x8.
297. Multiple Roots. If we apply Taylor's theorem succes
sively to/(ic) and/'(ic), we obtain
(9)/(») =
(10) /'(») =
f\a) +f"{a){x  a)+i^ {x  ay+^^ix  ay+ .....
XIX, § 297] GENERAL POLYNOMIAL FUNCTION 461
If /(a)= 0, the first relation shows that x—a is a factor of /(a;) ;
this constitutes a new proof of the factor theorem. If /(a;) 13
divisible by a; — a but not by {x — ay, it follows that /(a) =n d
and that f'{a) ^ 0. Hence by (10), or by the factor theor3n_,
fix) is not divisible by a; — a. If f{x) is divisible by (x — a)
but not by (x — df, we have, from (9), /(a)=0, /'(a)=0,
f"{a)=^0. We then conclude from (10) that if a is a double
root of f{x) — 0, it is a simple root of f'{x) = 0. In general, if
f(x) is divisible b}^ (x — a)* but not by (x— a)*+\ relation (9)
shows that /(a) =/'(«) =/''(«) = =/''(«) =0; /^a) =?t 0.
Hence, by (10), f'(x) is divisible by {x — a)*~^ but not by
(a? — a) *. This leads to the following theorem.
A simple root of f(x)= is not a root of f\x)=0. A double
root of f(x)= is a simple root of f'{x)= 0. In geyieral, a root
of order k off(x)= is a root of order k — 1 off'(x)= 0.
The following corollary of this theorem is evidently true.
Any multiple root off(x)==0 is also a root off\x)=0. If
f(x) andf'(x) have no common factor, f(x) = has no multiple
roots. If(f> {x) is the H. C. F. off(x) and f'{x), the roots of
<ft (x)=0 are the multiple roots off(x) = 0.
Example 1. Examine for multiple roots the equation
/(x) = x3 4 x2  10 x + 8 = 0.
We have /' {x) =Sx'^ + 2x 10. To find the H. C. F. of /(x) and f\x)
we proceed as in § 259 :
8a8f3a:230x+24
3x8 + 2x2 lOx
x + 1
3x
X2
20 X
+ 24
8x2
60 X
+ 72
8x2 +
2x
10
62X + 82
3x2
+ 2x
10
186x2
+ 124x
620
186x2
+ 246X
370 X
620
It is now clear that /(x) and /'(x) have no common factor,
we conclude that/(x)= has no multiple roots.
Hence
462
MATHEMATICAL ANALYSIS [XIX, § 297
Example 2. Examine for multiple roots the equation
/(x)=x*2x3 + 2xl = 0.
We have /'(x) = 4 x^  6 x^ + 2.
2x*
2x*
4x8 + 4x
3x8+ a;
2
 x8 + 3x
2x86x
2x83x2
2
+ 4
+ 1
3x26x
x22x
+ 3
+ 1
X
2x33x2 + 1
2x34x2 + 2x
x2 _ 2 X + 1
a;2 _ 2 X + 1
2x41
Hence (x — 1)2 is the H. C. F. of /(x) and /'(x), i.e. x = 1 is a triple root
of / (x) = 0. The fourth root of /(x) = is x = — 1. How is it obtained ?
EXERCISES
1. Examine for multiple roots each of the following equations :
(a) x33x224x28=0. (6) x* + xS + 1 = 0.
(c) x57x32x2 + 12x + 8 = 0.
Id) x5 + X*  9 x8  5 x2 + 16 X + 12 = 0.
(e) x46x8 + 12x210x + 3 = 0.
(/) x83x6 + 6x83x23x + 2 = 0.
2. Prove that the graph of y =/(x) is tangent to the xaxis at a point
representing a multiple root.
3. Prove that the graph of 2/=/(x) crosses or does not cross the
Xaxis at a point representing a multiple root according as the order of
the root is odd or even. [Hikt : Use Taylor's theorem.]
4. Prove that a root of order k of /(x) = is a simple root of
/*i(x)=0.
298. Complex Roots. If a \ hi (a, 6 real numbers, i2= — 1)
is a root of an algebraic equation /(«) = with real coefficients,
then a — bi is also a root of the same equation,
By hypothesis a f 6i is a root of the equation
f{x) = anx + a„_iaJ"^+ ••• + ao = 0,
i.e. f{a + 60 = a^ia + biY+ a„_i(a + bif^^ + ••• f Oo = 0.
If each of the terms in the preceding expression be expanded
XIX, § 298] GENERAL POLYNOMIAL FUNCTION 463
by the binomial theorem, the powers of i reduced to their
lowest terms (i^ = — 1, ^s = — i, etc.), and terms collected, we
obtain ^, , t\ n , r\'
where P represents the sum of the terms independent of i ajid
Q is the coefficient of L
But since P + Qi = hj hypothesis, it follows from § 281,
that both P = and Q = 0. We wish to prove that a — bi is
a root of f(x) = ; i.e. f{a — bi) = 0. To prove this we
merely have to notice that f(a — bi) may be obtained from
the expression for /(a 4 bi) by replacing i by — i. Therefore
f{abi)=P Qi,
where P and Q represent the same quantities as before. But
we have just shown that P = and Q = 0. Therefore
/(a — bi) = or a — 6i is a root oif(x) = 0.
EXERCISES
1. Solve X* + 4 a;3 + 5 a:2 + 2 X  2 = 0, one root being  1 + i.
2. Solve a!:* + 4a:S + 6a;2f4x + 5=:0, one root being i.
3. Solve a;4  2 x3 + 5 a;2 — 2 a: + 4 = 0, one root being 1 — i Vs".
4. If a+ Vb (a and b rational but V6 irrational) is a root off(x) =
with rational coefficients, a — V6 is also a root.
[Hint : Show that /(a + y/b) reduces to the form P + Qy/b where P
and Q contain only integral powers of b and Q is the coefficient of Vb.
Since P + QVb = 0, P = and Q = 0. Why ?]
5. Solve 2 x*  3 a;3  16 x2  3 X f 2 = 0, one root being 2 + VS.
6. Form an equation with rational coefficients, of which two of the
roots are i and 1 + \/2.
7. Solve the equation x^  (4 + V3)x'^ + (5 + 4 V3)x  5V3 = 0, if
one root is 2 — i.
8. Solve the equation x^ — (5 + i) x^ + (9 + 4i)x — 5 — 5 1 = if one
root is 1 + i. Is 1  I a root ?
464 MATHEMATICAL ANALYSIS [XIX, § 299
299. To Multiply the Roots of an Equation. To trans
form a given equation f(x)= into another whose roots are those
of /(a;) = each multiplied by some constant 7c, multiply the
second term of f{x) by Jc, tJie third term by k^, and so on, taking
account of the missing terms if there are any.
The required, equation is f(y/k) = 0. For, if f(x) vanishes
when X = a,f(y/k) will vanish when y = ka. Hence, if the given
equation is a„a;" f a^_^x''~^ + ... f ao = 0, the required equation is
""(!)" +""■©"""'■■■ +""=''
which on multiplication by k"" becomes
ttn?/'* H kanif"^ + k'^an2y''^ + • • • + k^ao = 0.
If A: = — 1, we have, the roots of f(^x) = are equal respec
tively to those off{x) = with their signs changed.
Example 1. Transform x^— ^x"^ { 5 = into an equation whose
roots are twice those of the given equation. The desired equation is
x^  4(2)a;2 + 6(2)3 = 0, or x^  8 x2 + 40 = 0.
Example 2. Transform x'' — Sx^\4iX^2x\l = into an equa
tion whose roots are those of this equation with their signs changed.
The result is (a:)73(x)5 + 4(x)42(x) + 1 = 0, or
a;7 _ 3 x5 4 x*  2 x  1 = 0.
EXERCISES
Obtain equations whose roots are equal to the roots of the following
equations multiplied by the numbers opposite.
1. x6  2 x^ + x + 1 = 0. (2) 3. x4  x2 + X + 1 = 0. (3)
2. x76x8+2x 1 = 0. (2) 4. x^ + x* x^ + x 1 = 0. (2)
Obtain equations whose roots are equal to the roots of the following
equations with their signs changed.
5. x7  6 x« + 2 X*  X + 1 = 0. 7. x7  x« + x6  X*  2 = 0.
6. xi6 _ 1 = 0. 8. 1  X  xa  x8  X*  x6 = 0.
XIX, § 300] GENERAL POLYNOMIAL FUNCTION 465
300. Variations in Sign. A variation of sign or change of
sign is said to occur in f(x) whenever a term follows one oi
opposite sign. Thus the equation o^ — 3 a;^ 47=0 has two
variations of sign.
If f(x) has real coefficients and is exactly divisible by x — 7c,
where k is positive, then the number of variations of sign in the
quotient Q{x) is at least one less than the number of variations of
sign infix).
Before proving this statement let us consider the process of
dividing /(a;) = x^{ x^ — 3 x^— 2 x^ — x^ + b x—lhj x — 1 and
/(x) = a;^ — a^f 4 a;2— 13 cc h 2 by a; — 2, making use of synthetic
division.
• 1 1 _3 _2 1 5 1 g.
1 2134 1
Q(a;) = l 2134 1
1 _1 4 13 2 2
2 2 12 2
Q(a!)=l 161
It will be noted in these examples that Q(x) has no varia
tions except such as occur in the corresponding or earlier
terms of f{x) and that since f(x) is exactly divisible by the
given divisor, the sign of the last term of Q{x) is opposite to
that in f{x). Let us now prove the statement in general.
Proof : From the nature of synthetic division it follows that
the coefficients in Q(x) must be positive at least until the first
negative coefficient of f{x) is reached. Then, or perhaps not
until later, does a coefficient of Q{x) become negative or zero,
and then they continue negative at least until a positive
coefficient in f{x) is reached. Therefore Q{x) has no variations
except such as occur in the corresponding or earlier terms of
f(x). But by hypothesis f(x) is exactly divisible by a;— A; and
2h
466 MATHEMATICAL ANALYSIS [XIX, § 300
hence the sign of the last term in Q(x) must be opposite to
that in f(x). Therefore the number of variations of sign in
Q(x) must be at least one less than the number of variations
of sign in /(a;).
301. Descartes's Rule of Signs. Theequationf{x)=0 with
real coefficients can have no more positive roots than there are
variations of sign in f(x) and can have no more negative roots
than there are variations of sign inf(— x).
Proof: Let r^, ^2, — , ^^(i? ^ ^) denote the positive roots of
f(x) — 0. If we divide f{x) by a; — ri, the quotient by ic — rj,
and so on until the final quotient Q{x) is obtained, then we
know from the last theorem that Q(x) contains at least p fewer
variations of sign than f{x). But the least number of varia
tions of sign that Q{x) can have is zero. Therefore f{x) must
have at least p variations, i.e. at least as many variations as
f{x) =0 has positive roots.
Second, by § 299, we know that the negative roots of /(a;)=0
are the positive roots of /(— a;) = and, hence, by the first
part of this proof, we know that their number cannot exceed
the number of variations of sign in/(— «).
It is important to notice that Descartes's rule of signs does not tell us
how many positive and how many negative roots an equation has. It
merely tells us that an equation cannot have more than a certain number
of positive roots, and cannot have more than a certain number of nega
tive roots.
Example. What conclusions regarding the roots of the equation
/gT _ 4 x5 + 3 a;2 — 2 = can be drawn from Descartes's rule ?
The signs of /(a;) are f 1 — , i.e. there are 3 variations and hence
the equation has no more than 3 positive roots.
The signs of /( — ic) are h + — , i.e. there are two variations and
hence the equation has no more than 2 negative roots.
But the equation is of degree 7 and has 7 roots. Therefore the equa
tion has at least two imaginary roots. Can there be more than two
imaginary roots ?
XIX, § 303J GENERAL POLYNOMIAL FUNCTION 467
EXERCISES
What conclusions regarding the roots of the following equations can
be drawn from Descartes' s rule ?
1. x^  2 x6 + X*  1 = 0. 4. x^  2 X* + x^  x^  X + 1=0.
2. x^ + X* — x^ + 1 = 0. 5. x**  1 = 0. (w odd)
3. x23  34 xi2 + X — 45 = 0. 6. x**  1 = 0. (n even)
7. Show that the equation x^ — 5x2 — x + 10 = has at least two
imaginary roots. How many may it have ?
8. Show that the equation x^ + x^ix — 1=0 has two and only two
imaginary roots.
9. Show that the equation x^ + 4 x' + 2 x — 10 = has six and only
six imaginary roots.
10. Can you tell the nature of the roots of the equation x* + ix^ — 3 ix
+ 4 = 0?
302. Equations in ^form. If each term of the equation
f{x) = a„a;" + a,,_,x^' + ... + a^ =
is divided by a„ (by hypothesis an ^ 0), we obtain the equation
x^ + p^x'''^ + PiX""^ + ... + jp^ = 0,
in which the leading coefiicient is unity and p^ = ^^, etc. An
equation in this form is said to be in the pform. For many
purposes this is the most convenient form.
303. Rational Roots. A rational root (^0) of the equation
f{x) = when the equation is in the pform with integral coeffi
cients is an integer and an exact divisor of the constant term.
Proof. Suppose that the equation f(x)= has a root a/b
where a/b {b > 1) is a rational fraction in its lowest terms.
Then we have
(11) (tf + p/tT' + ■■■+ pUt) +P, = 0.
I,  " \b
468 MATHEMATICAL ANALYSIS [XIX, § 303
Multiplying both, members of (11) by 6"~^ we have
or
(12) T^~ (^1^""' + ihci^'^h +  + Pnh^"),
The righthand member of (12) consists of terms each of
which is an integer. The lefthand member of (12) is a
fraction in its lowest terms. Therefore the assumption
that the fraction a/h is a root of f{x) = leads to an
absurdity.
Now suppose r (:^ 0) is an integral root. Then
rn hpir"! f p^r^^ + ••• h i?« = 0.
If we transpose the constant term pn and divide by r, we
obtain
(13) r^i^p^r^2+ ... .p^_^ = _P2.
r
Now each term of the lefthand member of (13) is an integer ;
hence pn/r must be an integer, i.e. pn must be exactly divisible
by r.
304. To Find the Rational Roots of an Equation with Ra
tional CoeiKcients. If the equation is not in the pform with
integral coefficients, reduce it to that form and then make use
of the results in § 303. The following examples will explain
the methods.
Example 1. Solve the equation x^ + S x^ — 4 x ^ 12 = 0.
By Descartes's rule of signs we know that the equation has no more
than one positive root and no more than two negative roots. From the
last article we know that if the equation has rational roots they are
factors of 12. Thus we need only try 1, — l, 2, — 2, 3,  3, 4, — 4, 6,
 6, 12, 12.
XIX, § 304] GENERAL POLYNOMIAL FUNCTION 469
By synthetic division we have
1 3412 12
2 10 12 I
16 6
The depressed equation * ia x^ \ ^ x + 6 = (x ^ S)(x + 2)= 0. There
fore the roots of the original equation are 2, — 3, — 2.
Example 2. Solve the equation 2a:8_^a;2 + 2x + l = 0.
Writing the equation in the )f orm we have
X3 + I X2 + X + ^ = 0.
If we multiply the roots of this equation by k, we obtain
x^\kx^\k'^x + — =0.
^ z
If we choose k equal to 2, this equation becomes
(14) x^\x^h4:X + 4 = 0,
an equation whose roots are twice those of the original equation.
By Descartes's rule of signs equation (14) has no positive roots. Any
rational roots are then negative, and are factors of 4, i.e. — 1, — 2, — 4.
By synthetic division
1 14 4 1
104
10 4
The depressed equation is a;^ + 4 _ q. Therefore the roots of (14) are
— 1, 2 1, — 2 I and the roots of the given equation are — ^, i, — i.
EXERCISES
Solve each of the following equations.
1. ic« + 5 x2 f 15 X + 18 = 0. 4. 6 a;3 + 7 x2 _ 9 x + 2 = 0.
2. xs + x2 + X + 1 = 0. 5. 6 x3  2 ic2 + 3 X  1 = 0.
3. x=i + x24x4 = 0. 6. 2x4 + 3x310x212x+8=0.
Find the rational roots of each of the following equations.
7. X*  3 x2  4 = 0. 10. 2 X*  x8  5 x2 + 7 X  6 = 0.
8. x6  32 = 0. 11. 2 X* + 2 x3  x2 + 1 = 0.
9. X* + x8 + x2 + X + 1 = 0. 12. 4 x*  23 x2  15 X + 9 = 0.
* If r is a root of a given equation /(.x*) = and f(x) = (x — r)Q(x), then the
equation Q{x)= is called the depressed equation.
470
MATHEMATICAL ANALYSIS [XIX, § 305
306. The Solution of an Equation with Numerical Coeffi
cients. The preceding articles furnish a number of methods
for attacking the problem of finding the roots of an algebraic
equation/ (a;) = with given numerical coefficients.
(1) We may examine the equation for multiple roots (§ 297).
(2) If the equation f{x) = has rational coefficients, we can
find all the rational roots by a finite number of trials.
(3) When any root a has been found, we may divide f(x) by
X — a and thus make the finding of the remaining roots depend
on an equation of lower degree (the depressed equation).
306. Irrational Roots. Graphical Approximation. In order
to compute approximately any one of the real irrational roots
of an equation/ (a;) = whose coefficients are real numbers, we
require first a rough approximation to the root which is to be
computed. The graph of y =f(x) is a powerful tool for this
purpose. An example will make the method clear.
Example. Locate approximately the real roots of the equation
f(x) = x6  ISx'^ + 2 a; + 5 = 0.
A table of corresponding values of x and
f(z) is as follows.
I^'
4^
7ko_
::::^:i::i::
:i:::?:i::d=:
::::4?:i::: i:
^ f
1)^^ uy
:E/:E:::E:E:E:
X 2  1
1
2 3
/(a;) 8311 6
5
 11 137
Fig. 262
Figure 252 exhibits a rough graph of this
function constructed from this table. We
conclude that a root of the equation lies
between — 1 and 0, another between and 1 ,
and a third between 2 and 3.
Moreover Descartes 's rule tells us that this
equation can have no more than two positive
roots and no more than one negative root,
since there are only two changes of sign in
f(x) and only one in/(— x).
We have therefore located all the real roots
of this equation.
XIX, § 307] GENERAL POLYNOMIAL FUNCTION 471
A more accurate construction in the neighborhood of one of these points
enables us to get a better approximation. For example, the values x = 2.2
and 2.3 give us respectively y = — 1.97 and 6.21. By drawing a smooth
~
T
4* .
^^
" ^: :
^^
rr
,t
20
2 Ir
2e^'
,../
'
^ i*
^^'
,''
**
, *
*'
—
*^^ 
 __

"

■
Fig. 253
curve through the three points corresponding to x = 2, 2.2, 2.3 (plotted
on a large scale, Fig. 253) we may estimate the root of f{%) = to be ap
proximately 2.23.
307. Newton's Method of Approximation. Having found a
first approximation to a root of an equation /(«)= 0, we may
secure a better approximation by a
method first suggested by Sir Isaac
Newton (16421727). In Fig. 254 let
GC represent the graph of y—f{x) in
the neighborhood of a root a? = a of the
equation. Let OM^ — x^ represent the
approximation to the root found ; let
M^Px = ^1= /(xi). Let the tangent to the
graph at Pi(iCi, y^ cut the a;axis in T.
The abscissa OT will then, in general, be a much closer
approximation to the desired root. The equation of the
tangent at Pj is
(15) 2^/(^i) = /'(^0(^a^i).
Placing 2/ = and solving for x we have
Fig, 254
(16)
052= OT=Xy
/fe)
472 MATHEMATICAL ANALYSIS [XIX, § 307
where x^ denotes our second approximation. We have then
(17) ^2 = JCi + ^1,
where the correction hi is given by
(18) K = f^.
Example. Find by Newton's method a better approximation to the
root X = 2.23 of the equation x^ — 13 x''^ + 2 ic4 5 = discussed in § 306,
f(x) = ic513x2 + 2a; + 5.
/'(x) = 5a:*26x + 2.
/(xi)=/(2.23)=0.039.»
/'(xi)=/'(2.23)=67.67.»
Hence we have
^~ /'(a^i) 67.67
whence X2 = 2.23057.
308. The Accuracy of Newton's Method. A question that nat
urally arises is : How accurate is this root, i.e. to how many decimal
places is it correct ? Taylor's theorem gives us information on this point.
We have ,^
(19) /(xi + ;iO = /W + /'(^i)'^i+'^^^^i'+ .
If our first approximation to the root is x = Xi and hi is the correction, f
Newton's method gives to ^i a value which makes the sum of the first two
terms of Taylor's expression vanish. Since h\ is very small, the terms
beyond the third (involving h\^ and higher powers of ^i) are insignificant
fii (xA
compared with the term , . h^. Hence for our purpose we may write
2i I
(20) /(a:i + Ai) = i/"(a;i)A;2.
In the example considered above we have
/"(x)= 20x326. /"(a^i) =/"(2.23)= 195.8.
hi^ = (0.00057)2 = 0.00000032.
Hence we have
\f"{x{)hi^ =/(xi + h{) = 0.0000313.
♦ Use synthetic division to get these values.
t In the example just cousidered ^i = 0.00057.
XIX, § 309] GENERAL POLYNOMIAL FUNCTION 473
Moreover ^,^^^ ^ ^^^ ^^,^^^^ +r{x{)h, + ..,
and in this example /'(aii + hi) = 67.67 + 0.11 = 67.78 approximately. It
follows that the new correction is about
;i2 = ^/(?l+M<_ 0.000001.
f'{xi + hi)^
Therefore we may conclude that x = 2.23057 is the root sought, to five
decimal places.
EXERCISES
Find to three places of decimals the irrational roots of the following
equations.
1. a;3 + 3x + 20=0. 2. a;^ + 2x2  a; + 3 = 0. 3. a^ + a;l = 0.
4. a:8 + 4 a;2 _ 6 = 0. 5. x^ + Sx"^  Sx1 = 0.
6. If X is the cosine of an angle and y is the cosine of one third of the
angle, then 4 y^ = 3 ?/ + x. Find the value of cosine of 20° to three places
of decimals.
7. An open box is to be made from a rectangular piece of tin 9 x 10
inches, by cutting out equal squares from the corners and turning up the
sides. How large should these squares be so that the box shall contain 59
cu. in,?
8. Find the cube root of 12 ; 45 ;  37.
309. The Relation between Roots and Coefficients. If
fit ^2) •••? ^„ are the roots of the equation x"" + picc""^ 4 P2^"~2 __
— hPn = 0, then
x^ { PiX""^ \ PiX""^ + ... \p„=(xr{){xr2){xr,;).
If we carry out the indicated multiplication in the righthand
member and equate the coefficients of like powers of x, we '
have
(21) Pi = rir2 r,,
(22) JP2 = nrz + nra + ... + nr^^ + r^n +  + »*niV
(23) Ps =  ^1^2?*3  r^r^U ^n2^nl^n.
(24) P«=(l)Vir,...r,.
474 MATHEMATICAL ANALYSIS [XIX, § 309
That is,
— Pi= the sum of the roots.
JP2 = the sum of the products of the roots taken two at a time.
— Pg = the sum of the products of the roots taken three at a time.
(— lyPn = the product of all the roots.
We have at once the following corollaries :
1. To transform an equation into another whose roots are those of the
original equation each midtiplied by m, multiply p\ by m, p2 by m^, p^ by
m^, and so on (§ 299).
2. To transform an equation into another whose roots are equal to
those of the original equation with their signs changed, change the signs
of the alternate terms, beginning with the second.
Example 1. Solve the equation 2y:^— x'^ — 9>x + ^ = Q given that
two of the roots are equal in absolute value but opposite in sign.
Let the roots be r, — r, and s.
Then r — r + s = \,
rs — rs — r^ = — 4:,
r2s=2.
Therefore s = ^ and r = 2 or — 2, i.e. the roots are ^, 2, — 2.
EXERCISES
1. Solve a* + x* — 4 5c — 4 = 0, given that the sum of two of the roots
is zero.
2. Solve x^ — 6 a;3 — 9 x2 + 54 x = 0, given that the roots are in arith
metic progression.
3. Solve X* 16 x3+ 86 x^ 176 x + 105 = 0, given that the sum of two
roots is 4. Ans. 1, 3, 5, 7.
4. Solve 4 x8 — 20 x2 — 23 X — 6 = 0, two of the roots being equal.
5. If n, ro, rs are the roots of x^ — 5 «'* + 4 x — 3 = 0, find the value
of each of the following expressions :
(a) ri2 + rz^ + n^.
(6) ri^ \ r2^ + ra,^
(c) riV + ri^rs^ + r^^rs^.
(d) ri^ri + ri^r^ + r2'Vi + rz'^rs + rs^i + r^^r^.
CHAPTER XX
DETERMINANTS
310. Determinants of the Second Order. Expressions of
the form aib2 — a^bi , where a^, tt2, 61 , 62 ^'^^ ^^J numbers, arise
often in mathematical analysis. Thus the area of a triangle
with one vertex at the origin and the other two vertices at the
points (ai , bi), (a2 , 62)? is equal to ^{aib^ — a26i) (§ 195). Again,
the solution of a pair of simultaneous linear equations in two
unknowns (§ 69) can be written as two fractions whose numer
ators and denominators are all of this form. (§ 311.)
The expression aib^ — ^2^1 ^^J ^^ written in the form
I ai bA
and is then called a determinant of the second order. Such a
determinant contains two rows and two columns. The numbers
«i > «2 J ^1 J ^2 J are called the elements of the determihant. The
two elements ai, 62 form the socalled principal diagonal.
To evaluate a determinant of the second order, i.e. to find
what number it represents, one merely has to subtract from
the product of the terms in the principal diagonal the product
of the other two terms. Thus we may write
tti bi
0"2. ^2
= aib^ — a^bi ;
4 7
3 6
= (4)(6)(3)(7)=45.
It is important to notice that each term of the expansion
contains one and only one element from each row and one and
only one element from each column.
475
476
MATHEMATICAL ANALYSIS [XX, § 311
311. Simultaneous Equations in Two Unknowns. Let the
equations be
(2)
1 OgflJ + 62S/ = C2.
If we solve these equations by the usual method of elimination
(§ 69), we obtain
(3)
C162 ~ ^2^1 ^l^ — ^2^
"" aib2 — a^hi ' "" 0162 — a2&i '
provided a^h^—a^hi^O. We at once recognize the fact that
these results may be written in the form
(4)
x =
Ci 61
ai Ci
C2 62
02 C2
«! h
'} y —
«! &1
^2 &2
02 &2
provided 0162 — c^h =^ 0. The following points should be noted
in the above solution.
(1) The determinants in the denominators are identical and
are formed from the coefficients of x and y in the original
equations. .
(2) Each determinant in the numerator is formed from
the determinant in the denominator by replacing by the
constant terms the coefficients of the unknown whose value
is sought.
Example. Solve by determinants the simultaneous equations
(2xy = l,
[Sz + 2y = S.
Solution :
1
1
2
1
3
2
6
= 7' y=
3
3
2
1
2
1
8
2
3
2
XX, § 311]
DETERMINANTS
477
EXERCISES
Evaluate each of the following determinants
^ 14 61 ^ I — sin a — cos a I
1 • 3.
1 3 1 1 I cos a sin a I
12 a 61 Itan^ sec^l
c — d\ ' sec ^ tan d
I sin e cos 6
sin a cos a
2.
6. Show that the normal form of the equation of a straight line
(§ 205), may be written in the form
.* " Up.
— sm a cos a \
Solve by the use of determinants the following pairs of equations :
f2x + y = 3, g
\ 6x^ y = 4.
Ax Sy_
= 2,
¥— •
9.
x + y = l,
3 *^ 3
10.
11.
ic sin ^ 4 y cos ^ = sin 0^
xcosd ■{■yemd = cos 6.
X \ y tan ^ = sec^ 6,
X sec2 d +yctne = sec* ^ + 1
^ns. 1, tan e.
Prove the following identities and state in words what they show.
12.
13.
14.
15.
16.
ai &i
Ol
02
0,2 62
61
62
dl Oi
02 O2
=
0.
ai 61
a2 62
mai ?)i
maa 62
(ai + 5i) 61
(a2 + &2) &2
= r»
61 ai
62 02
«i &i
02 62
aa 62
* For example, Ex. 12 shows that in a secondorder determinant if the cor
responding rows and columns are interchanged, the value of the determinant
is not changed.
478
MATHEMATICAL ANALYSIS [XX, § 312
312. Determinants of the Third Order. To the square
array
bi Ci
(5)
O2 C2
^3 Cs
we assign the value
(6) aidgCg + azhci + ag^iCz — 016302 — aa^iCg — a^bzCi
and the name determinant of the third order.
The expression (6) is known as the development or expan
sion of the determinant, the numbers ai, hi, etc., as the elements,
and the elements Oi, 62, Cg as the principal diagonal.
It is important to notice that in the development (6) each
term consists of the product of three elements, one and only
one from each row and one and only one from each column.
An easy way of obtaining the expansion (6) of the deter
minant (5) is as follows :
Form the product of each element of the first column by the
secondorder determinant formed by suppressing both the row
and column to which the element belongs. Change the sign of
the product which contains the element in the first column and
the second row and take the algebraic sum of the three products.
Example 1.
Example 2.
Example 3.
a\ 61 ci
62 Co
hx Ci
&1 Ci
ai 62 C2
= a\
a'i
+ as
hz ca
hz cs
62 C2
az 63 C3
ai&2C3 — ai&3C2 — «2^iC3 + «2&3Ci + a^biCi — azbiCi.
2 3
6 4
4 1
3 1
1
6
+ 6
il + ^U
= 2(4 + 7)H 6(3 + 2) + 4(21  8)= 99.
= 3
= 3(_220):
m.
XX, § 312J
DETERMINANTS
479
EXERCISES
Evaluate each of the following determinants.
1.
2
1
3
1
1
1
1
1
2
2.
1
1
5
2
2
2
. 3.
6
3
3
5
1
7
4
41
la 6 c
6 <r c
c a 6
6. In § 196 it was shown that the area of the triangle whose vertices
are Pi(a:i, yi), P2(2C2, 2/2)1 ^3(^:3, tjs) is
^[Xi?/2  X2?/l + 3^22/3  X3?/2 + Xa^l  Xi^s]
Prove that the area of this triangle is
xi yi 1
I X2 y2 I ■
X3 VZ 1
6. Using the result of Ex. 5, find the area of the triangle whose
vertices are
(a) (2, 1), (3, !),(!, 7);
(6) (3,2), (3, 6), (1,4);
(c) (0, a),.(0, a), (&, 0).
7. Prove that the three points Pi(xi, ?/i), P2(a;2, 2/2), P3(a53» ys) are
coUinear if, and only if.
a^i yi
1
X2 y2
1
X3 yz
1
8. By means of determinants show that the three points (a, & + c),
(&, c \ a), (c^ a + b) are collinear.
9. By use of determinants determine whether the three points (0, 0),
(1, 1), (5, 6) are collinear.
10. Prove that the equation of the straight line through the points
Pi(a:i, t/i), P2(X2, 2/2), is
= 0.
X
y
1
Xi
yi
1
X2
2/2
1
11. By determinants find the equation of the straight line through each
of the following pairs of points!
(a) (2, 1), (3, 7) ; (6) (6, 1), (2,  1); (c) (7, 1), (9, 1).
12. Find by the use of determinants whether the three lines 3 a;— y — 7
= 0, 2x4j/ + 2=:0, a: — y = are concurrent or not.
480
MATHEMATICAL ANALYSIS [XX, § 313
equations be
(7)
313. Solution of Three Simultaneous Equations. Let tha
aiOJ 4 6i2/ + CiZ = dj,
If we solve these three simultaneous equations by the usual
method of elimination, we obtain,
dibzC^ + dzbsCi + dsbiCz — dzb2Ci — dih^^Cz — d^bic^
(8)
~~ ciiboCs + ota^gCi + a^biCz — a^bzCi — aib^Cz — azbic^
_ aid2Cs + a2C?3Ci + a^diC2 — a^doC} — ai(?3C2 — aodiC^
z =
«]&2C?3 + «2^3C?1 + «3^1<^2 — «3^2f'l
«i&2C3 + (hbsCi \ (136102 — 0362^1
O163CZ2 — OL^id^
^1
&1
Ol
C?2
62
Ci
C^3
&3
C3
«1
bi
Cl
tti
b.
C2
<h
h
C3
cti
^1
Cl
<H
(i2
C2
V —r.
as
d.
C3
if
«!
bi
Cl
a2
&2
C2
^3
^^3
C3
ai
&i
C?i
ag
&2
d,
^3
63
C?3
ai
^^1
Cl
ag
60
C2
03
&3
C3
provided the denominator of each fraction is not zero. These
results may be written in the form
(9)
Each denominator is the same determinant, which is called
the determinant of the system. It is made up of the coefficients
of X, tfj z. Each determinant in the numerator is formed from
the determinant in the denominator by replacing the coefficients
of the unknown whose value is sought by the constant terms.
Compare this rule with that given in § 311.
ExAMPLB. Solve the following equations by determinants :
f6x22=2,
3y4;2 = 7,
2x6j/ = 10.
XX, § 313]
DETERMINANTS
481
Solution.
2 21
7 3 4
195 ol
u
2
3
4
5
224
112
= 2;2/ =
5
2
2
7
4
2
19
336
5
2
 112
3
4
2
6
5
2
3
7
2
5
19
5
2
3
4
2
— 5
448
112
3.
EXERCISES
Expand each of the following determinants :
4.
Solve by determinants each of the following sets of equations ;
2 3
4 1
1 4
3
2
1
1 1 1
a h c
a2 62 c2
7 1
2 2
4 2
2
► 6
4
a h g
h b f
g f c
a;
z
z
a;
y
y
y
z
X
[4aj45y + 22r = 20,
6. j 3 X  3 ?/ + 5 5! = 12,
[5x + 22/40=3.
Ans. (1, 2, 3).
[x^y ■\z=\,
8. ax f 6y + cs = d,
[ cC^x + 6'^?/ 4 cz = d^.
10. Solve the equation
7.
3x + yz = 3,
x + y + z =7,
2x + 4y \z=12.
ax + y — z = a'^ + a — l,
— x + ay\z = a^— a\l,
x — y + az = a.
Ans. (a, a, 1),
1
1
5
0.
11. Solve for x and y the simultaneous equations
2i
x+1
3
y
2 1
X 1
2 1
= 0,
X
2
y
0.
482
MATHEMATICAL ANALYSIS [XX, § 313
12. Evaluate the determinant
sin a cos /S 1
cos a sin /3 1
1 1 1
Prove the following identities and express in words what they prove.
See Ex. 1216, pp. 477.
13.
15.
17.
«! &i Ci
a^ hi C2
az h C3
«! tti &i
Gi a^ bi
as «3 &3
ai
tti «3
ai 61 Cl
bi
bz 62
. 14.
a2 62 C2
= 
Cl
C2 C3
^3 bs C3
ai
bi
Cl
as
bz
C3
a2
62
C2
0.
16.
max 61 Cl
ma2 62 C2
maz bz cz
ax
61
Cl
as
&2
02
as
68
cs
(ai + 61) 61 Cl
(«2 + 62) 62 C2
(as + bz) bz C3
ai
bx Cl
a2
62 C2
az
bz cz
314. Inversions. Let us consider the permutations of a set of ob
jects, such as letters or numbers, and let us fix a certain particular order
of the objects which we shall designate as the normal order. An inversion
is said to occur in any permutation when an object is followed by one
which in the normal order precedes it. Thus if abed is the normal order,
then there are two inversions in bade. If 1234 is the normal order, then
there are three inversions in 1432.
Theorem. If in a given permutation, two objects are interchanged,
the number of inversions with respect to the normal order is increased or
decreased by an odd number.
Let us consider the permutations Xrs Y and Xsr Y, where X and ;i"
denote the groups of objects which precede and follow the interchanged
objects r and s. Any inversion in Xand Fand any inversi(m due to the
fact that X, r, s precede Y are common to Xrs Y and Xsr Y. Therefore, the
number of inversions in Xrs Y is equal to the number in Xsr Y increased
or decreased by 1 (according as rs is or is not in the normal order).
Now let us consider two objects such as r and s separated by i objects.
If the objects r and s are interchanged, the number of inversions is still
changed by an odd number. For, by f + 1 interchanges of adjacent pairs
the object r can be brought into the position immediately following s, and
by i further interchanges of adjacent pairs, s may be brought to occupy
XX, § 316]
DETERMINANTS
483
the position formerly held by r. Each of these (i f 1) + i = 2 i 4 1
interchanges of adjacent pairs has increased or decreased the number of
inversions by 1. Hence the net result of these 2^ + 1 interchanges has
increased or decreased the number of inversions by an odd number.
315. Determinants of the nth Order. The square array
(10)
ai 6i
a2 62
an K
of n^ elements, such as we have considered for the cases n = 2, w = 3, is
called a determinant of the nth order and will be denoted by the Greek letter
A. This determinant will he understood to stand for the algebraic sum of
all the different products of n factors each that can be formed by taking
one and only one element from each row and one and only one element
from each column, and giving to each such product a positive or negative
sign according as the number of inversions of the subscripts {normal
order 1, 2, •••, n) is even or odd, when the letters have the normal order
ab 'q.
It should be noted that from the remarks in § 314 it follows that if we
arrange the elements in any product so that the subscripts are in normal
order, we can determine the sign of each term, by making it positive or
negative according as the number of inversions of the letters is even or odd.
316. Properties of Determinants. Theorem 1. The expan
sion of a determinant of order n contains n ! terms.
Proof. There are as many terms in the expansion of a determinant
of the nth order as there are pennutations of the subscripts 1, 2, 3,, n.
But this number is n ! (§ 269) .
Theorem 2. If each element of any row or column is multiplied by any
constant m, the value of the determinant is multiplied by m.
Proof. Since by the definition of a determinant, each term of the ex
pansion must contain one and only one element from each row and each
.column, the factor m will appear once and only once in each term of the
expansion. If m is factored out of this expansion, the remaining factor
is the expansion of the original determinant.
484 MATHEMATICAL ANALYSIS [XX, § 316
Illustration.
inaib2C3\ma2b3Ci+mazbiC2—maibzC2 — mazbiCz — mazbiPi
moi
bi
Cl
mctz
b2
C2
= mt
maz
bz
cz
= m
ai bi
Cl
02 bz
C2
az bz
cz
Theorem 3. The value of a determinant is not changed if roics and
columns are interchanged, so that the first row becomes the first column,
the second row the second column, and so on.
This follows at once from the definition of the determinant and the
paragraph immediately following it (§ 815) .
Theorem 4. If two rows or two columns of a determinant are inter
changed, the sign of the determinant is changed.
Illustration. See Ex. 14, p. 477, and Ex. 14, p. 482.
Proof : Since by Theorem 3 rows and columns may be interchanged
without affecting the value of the determinant, we need only consider the
interchange of two rows. First, if two adjacent rows are interchanged,
the order of the letters in the principal diagonal and in each term of the
development is left unchanged. However two adjacent subscripts in each
term of the expansion are interchanged, and hence the sign of every term
is changed. Why ?
Next consider the effect of interchanging two rows separated by k inter
mediate rows. By k interchanges of adjacent rows, the lower row can be
brought just below the upper one. Now the upper row can be brought
into the original position of the lower row Toy k \ 1 further interchanges
of adjacent rows. Therefore interchanging the two rows is equivalent to
2k\l interchanges of adjacent rows. But 2 A; + 1 is an odd number and
therefore this process changes the sign of the determinant.
Theorem 5. If two rows or two columns of a determinant are identical,
the value of the determinant is zero.
Proof : Let A be the value of the determinant and let the two identi
cal rows or columns be interchanged. Then, by Theorem 4, the value of
the resulting determinant is — A. But since the rows or columns which
were interchanged were identical, the value of the determinant is left
unchanged. That is to say, A = — A or 2 A = 0, or A = 0.
Corollary. If all the elements in any row or column are the same
multiples of the corresponding elements in any other row or column, then
the value of the determinant is zero.
XX, § 317]
DETERMINANTS
485
317. Minors. If we suppress the row and the column in which any
given element appears, the determinant formed by the remaining elements
is called the minor of that element.
Illustration. In the determinant
the minor of az is
ax
6i
Cl
az
&2
C2
«3
&3
cz
and the minor of Cs is
The minor of a\ is denoted by A\, of hj by ^y, etc.
hi
Cl
bs
Cs
«i
bi
Cl2
62
EXERCISES
1. Prove that
2 2 3
1 1 5
4 4 9
=
0.
2.
Prove that
3. Prove that
4 5 6
2 1 5
1 5 3
=
4 2 1
5 1 5
6 5 3
•
4. Prove that
3 4 5
2 4 1
8 4 5
' = 
4 3 5
4 2 1
4 8 5
•
6. Prove that
4 1
3 1
2 2
1 3
5 4
6 5 3
5 1 3
23 6
3 2 9
1 1 12
= 0.
6. Prove that
26 9
28 18
30 3
<
5
10
25
= 30
13
14
15
3 1
6 2
1 5
3 5 8 1
1 2 5 =0.
2 4 10
7. How many inversions are there in the arrangement 4213765 if the
normal order is 1234567 ?
8. How many inversions are there in the arrangement 46321 if the
normal order is 42316 ?
486
MATHEMATICAL ANALYSIS [XX, § 317
9. Find the value of the minor of 5, of 6, of 7, for the determinant
4 5 1
3 6 2.
2 7 8
10. Write down the minor of as, of C2, of 64, for the determinant
[1 &i ci di
I2 62 C2 di
64 C4 d^
11. Show that
1 2 51
3 3 6 2
4 2 7 3
5 15 4
1
6 4 3
2
5
12 3
5 7 6
= 
1
4 3 2
2 12 3
15 4 3
14 3 2
5 5 7 6
318. Additional Theorems. — The following theorems will be
found useful in evaluating determinants.
Theorem 6. Laplace's Expansion. If the product of each element
in any row or column by its corresponding minor be given a positive or
negative sign according as the sum of the number of the row and the num
ber of the column containing the element is even or odd, then the algebraic
sum of these products is the value of the determinant.
Proof : First, it is evident that in the development of the determinant,
Ai is the coefficient of ai. For Ai is a determinant of order n — 1 in the
elements a2, •••, a„, and its expansion contains a term for each permuta
tion of 2, 3, ••, n. Moreover, the signs of the terms are correct ; for, the
number of inversions is not changed by prefixing ai.
Second, let us consider the element e situated in the zth row and the jth
column. We can bring this element to the leading position, i.e. first row
and first column, by i — 1 transpositions of rows and j — 1 transpositions
of columns, i.e. hji\j — 2 transpositions in all. Therefore the sign of
the determinant will have been changed i + j — 2 times. That is, if i + j
is an even number, the sign of the determinant is left unchanged ; while if
t+jis an odd number, the sign of the determinant is changed. Now
that the element under consideration is in the leading position, we know
from the first step that its coefficient is its minor. Since the relative posi
tions of the elements not in the ith row or the jth column are not effected
by these transpositions, the minor of the element in its original position
is the same as the minor of the element when it is in the leading position.
XX, § 318]
DETERMINANTS
487
Hence the coefficient of the element e, which is situated in the ith. row
and the jth column, is (— ly+J E, where E is the minor of the element e.
Corollary, If in the development of a determinant by minors with
respect to a certain column (row) the elements of this column {row) are
replaced by the corresponding elements of some other column {or row),
the resulting expression vanishes.
= aiAi — azAi + dsAs — a^A^i.
We wish to show that, for example, 61^1 — 62^2 + 63^3 — b^Ai is
zero. This expression is zero, for we have replaced the column of a's by
the column of ?)'s and hence the determinant has two columns identical.
The same proof applies to a determinant of order n.
Theorem 7. If each of the elements of any row or column of a deter
minant consists of the sum of two numbers, the determinant may be
expressed as the sum of two determinants.
Proof : Let
Illustration.
ai
bi
Cl
di
a2
62
C2
d2
as
63
ca
da
a4
64
C4
d.
(ai + a'l)
(a2 + a'2)
{an + a'„) 6,
Qn
be the given determinant. Expanding in terms of the first column we
have
(ai + a'i)Ai  (aa + ^'2)^2 + (aa + a'3)^3 +••• + ( 1)'*K«» + «'n)^n
or [aiAi — a^Ai + azA^ + ••• + (— l)"ia„A]
+ [a'1^1  a'2^2 + a'3^3 + ••• + ( l)"ia'„J„],
«1
bi
■ qi
02
62
■';
+
an
bn
•Qn
a'l
Theorem 8. If to the elements in any row {or column) be added the
corresponding elements of any other row {or column) each multiplied by
a given number m, the value of the determinant is unchanged.
The proof of this theorem follows easily from Theorems 7, 6, and 2.
488
MATHEMATICAL ANALYSIS [XX, § 319
319. The Evaluation of Determinants. We are now in a posi
tion to expand a determinant of any order. The following examples will
illustrate the methods employed.
Example 1. Expand
A =
Multiply the first column by
columns. It gives
A =
27
28
29
1 and add it to the second and third
25
26
26
27
27
28
25
1
2
26
1
2
27
1
2
By the corollary of Theorem 5, the value of this determinant is 0.
Example 2. Expand the determinant
216
A =
1
14 6 3
4 2 7 4
3 12 5
We seek to transform this determinant in such a way as to make all
the elements but one in some row or column 0. The second column
looks most promising. We accordingly add 4 times the first row to the
second row (this replaces the 4 in the second row by 0); we then add 2
times the first row to the third row (Why?) ; and then add the first
row to the fourth row (Why ?) These operations give
A =
21 5 1
9 26 7
8 17 6
5 7 6
9 26 7
—
8 17 6
5 7 6
—
2 26 7
2 17 6
1 7 6
The last determinant may be still further simplified as follows ;
A =
2 26 7
2 17 6
17 6
__ 140 191
"131 18
= _ (16231):
40 19
31 18
17 6
9 1
31 18
131.
* This determinant is obtained from the preceding by subtracting the
elements of the last column from those of the first.
XX, § 319]
DETERMINANTS
489
EXERCISES
Evaluate the following determinants.
5.
14
13
—
121
17
16
17
.
25
24
18
34
23
12
23
34
21
.
14
35
26
18
26
24
29
39
49
,
37
35
11
2 
2
1 1
1 
1
4 2
2 
2
1 1
•
2
1 1
3
4
2 6
4 
3
8 4
2
8
3
1
4 1
23
24
25 26
12
13
14 15
32
33
34 35
2
2
2
2
7.
8.
b c + d
c b + d
d 6 + c
2 a a^
a + b ab
2b 62
abed
6x00
c y
d z
10.
11.
1
1
1
a
6
c
.
a2
62
C2
1
1
1
1
a
6
c
d
a2
62
C2
(22
a8
68
C«
(28
12. abed
a 6 c (2
a — 6 c 5
a — 6 — c (2
13. Prove that if a determinant v^hose elements are rational integral
functions of some variable, as y, vanishes when y = 6, then y — b is a
factor of the determinant.
[Hint : Use the corollary of theorem 6.]
14. Solve by factoring Examples 10, 11, 12.
15. Factor into two factors
a b c
b c a
cab
16. Factor
a
a2
be
b
62
ca
c
C2
ab
490
MATHEMATICAL ANALYSIS
[XX, § 320
320. Solution of a System of Linear Equations. Suppose we
have n linear equations in n unknowns and we desire their solution. Let
the equations be
(11)
aixi H 6ia;2 + Cia3 +
a^Xi f biXi + C2Xs +
+ P\Xn = gi
QnXl + \X2 + CnXs + ••• + PnXn= ^n
Let A be the determinant of the cofficients of the unknowns, i.e.
(12)
A=
ax
6i •
.. px
at
62 .
•• Vi
an
K •
•• Vn
The determinant A is called the determinant of the system. Multiply
the equations by ^1, — J.2, Az, — A4, etc., respectively, and add the re
sults. Then we have
(13) Xi(aiAi — a^Az •••) + X2(hiA\ — h^Ai •••)+ ••
+ ic„(pi^ip2^2)
= qiAi—qzAi ••••
From the corollary of Theorem 6 it follows that the coeflBcient of x\ is
A and that the coefficients of the other unknowns are zero. Moreover,
the righthand member of (13) is the expansion of A if we replace the
column of a's by the column of constant terms. This determinant will
be denoted by A^^. Therefore we may write
or
provided A rjt 0.
Similarly
provided A :?t 0.
A . iCi = Ao
A
Xi
^
It will be noticed that this is a direct extension of the methods employed
in §§ 311, 313. The result may be stated in words as follows. The value of
any unknown is equal to a fraction whose denominator is the determinant
of the system and whose numerator is the determinant obtained from the
former by replacing the coefficients of the unknown sought by the column
of constant terms.
XX, § 322]
DETERMINANTS
491
321. The Case A = 0. The previous methods show that, even if
A = 0, we can derive from the given equations the relations
A . a:i = Aog, A • X2 = A^g, •••, A • a:« = Apq.
Now if A = 0, th6se relations would imply that
Aa3 = 0, A5, = 0, ..., Ap5 = 0.
But it is easy to write down a system in which A = and one or more of
the Aog, Aftg««are not zero. Such a system is then clearly inconsistent
and has no solution. For example, 2 xi + X2 = 1, 2 Xi + a;2 = 2.
If Aag = Afeg = ••• = Apg = 0, the system may be consistent but the un
knowns Xi, iC2, •••, Xn are not then completely determined. For example,
2 xi + 3^2 = 1, 4 jci + 2 X2 = 2.
A complete discussion of this case is beyond the scope of this book.*
322. Consistent Equations. Equations which have a common
solution are called consistent. Consider the three equations in two un
knowns X and y :
(14) aix + biy  ci = 0.
(15) a2X + hiy + C2 = 0.
(16) azx + hzy + Cg = 0.
Two cases arise according as to whether a pair of the three equations
has a single or an infinite number of solutions.
Case 1. A single solution. In order that these three equations be con
sistent it is necessary that
Cl
h
C2
&2
dl
&1
a2
&2
(17)
satisfy equation (16), i.e. that
or its equivalent
ai
C\
«2
C'2
ai
bi
a2
&2
La2 62I J
Cl bi
Ol Cl
ai 61
 as
63
+ C3
C2 62
(H C2
aa &2
=
= 0.
* Those interested in this problem will find a complete discussion in
BocHEB, Higher Algebra, Chapter IV.
492
MATHEMATICAL ANALYSIS [XX, § 322
Case 2. An infinite number of solutions. In this caae
ttl _ (Z2 __ Cli
bi 62 bs
and hence, by the corollary of theorem 5, the above determinant must equal
zero. Therefore, in order that three linear equations in two unknowns
have a common solution, it is necessary that the determinant of the coeffi
cients of the unknowns and the known terms vanish.
Extending this result to n linear equations in n — 1 unknowns, we
have a necessary condition that n linear equations in n — 1 unknowns he
consistent is that the determinant formed from the coefficients of the un
knowns and the knoion terms must vanish.
It must be clearly understood that the vanishing of the above determi
nant is only a necessary and not a suificient condition that the equations
be consistent. For example, the system
2x+ yl = 0,
2x+ y + 5 = 0,
4a; + 2y + 3 = 0,
gives
2 11
A= 2 1 6 =0,
4 2 3
but the equations are inconsistent, for any pair are inconsistent.
EXERCISES
Solve the following systems of equations by means of determinants ;
1.
8. <
2x — y — z = 0j
Zx + y+z = 6,
2xSy — v=—2,
2xhSv — d.
— a; + ?/ + 2 = 2m,
xy + z=2n,
x\y — z2p.
32/ 4x 20 + 10 =21,
a; + 72/ + 2w> = 13,
l/2a;30 + 2io = 14,
.3a; + 5^  50 +3wj = 11.
4.
x + y + w = Q,
x + y + = 7,
y + + w = 8,
X + + to = 9.
05 + 2^0 + 3to= 10,
x + 3y — 20 — 4to = l,
2xy30 + 5to = 3,
3x.y02io = 18.
XX, § 322]
DETERMINANTS
493
Determine whether the following systems of equations are consistent :
6. a; + y2 = 0, 7. ISx2y + 'l=0. 8. lxy\7=0,
[4x+y2 = 0. [3x + y2 = 0.
3a;6y2=0.
9. Find k so that the following equations are consistent :
2x + y3 = 0,
3ic2/ = 2,
x^y + k = 0.
/
MISCELLANEOUS EXERf^ISES
1. Prove that the equation of the oifcle that passes through the points
(aJi,yi), (2:2,2/2), (353,2/3) is
(x2 + y2) X y \
(ici^ + yi^) a:i yi 1
(X2^ + y'J^) ^2 2/2 1
(a;82 + 2/3^) a;3 2/3 1
= 0.
2. Prove that ax^ +
two linear functions if
+ 2 hxy + 2fx + 2gy + c is the product of
a h g
h b f
9 f c
3. Prove that a necessary condition that the three lines aix 4 biy + Ci
0, a2X + 622/ + C2 = 0, asx + 632/ + ca = 0, be concurrent is that
ai 61 Ci
052 62 C2 = 0.
as 63 C3
Is this condition also suflBcient ?
4. Prove that the locus of the equation ax^by + = is a straight
line.
[Hint : Let {xi ,2/1), (icz , 2/2) be any two fixed points on the locus and
{x, y) any other point on the locus. Then we have axi + 6yi + c = 0,
ax2 + by2 h c = 0, ax + by \ c = 0. Since these equations are consistent,
the determinant of the coefficient is zero.]
PART V. FUNCTIONS OF TWO VARIABLES
SOLID ANALYTIC GEOMETRY
CHAPTER XXI
LINEAR FUNCTIONS
THE PLANE AND STRAIGHT LINE
323. Introduction. Thus far the only functions which we
have represented geometrically are those of the form y = f{x),
i.e. functions of a single independent variable x. Such func
tions, in general, were seen to represent a curve in the (a;, y)
plane. We shall now study functions of the form z = f(x, y),
i.e. functions of two independent variables x and y. In order
to carry out this investigation it is necessary to set up a coordi
nate system in three dimensions.
324. Orthogonal Projections. The orthogonal projection of
a point P upon a plane a (Fig. 255) is the foot P' of the per
pendicular drawn from P to a. The ortho
gonal projection of a segment PQ upon a is
the segment PQ' joining the projections of
P and Q upon a.
The orthogonal projection of a point P
upon a line I is the foot P of the perpen
dicular drawn from P to I. The orthogonal projection of a
segment PQ upon the line I is the segment PQ' joining the
projections of P and Q upon I.
494
XXI, § 325]
^
LINEAR FUNCTIONS
495
325. Rectangular Coordinates in Space. Consider three
mutually perpendicular planes intersecting in the lines X'X,
F' F, ZiZ. These lines are themselves mutually perpendicular.
The three planes are known as the coordinate planes and their
three lines of intersection as the coordinate axes. The planes
are known as the xyplane, yzplane, xzplane, and the axes
as the Xaxis, yaxis, zaxis. The point which is common
to the three planes and also to the three axes, is called the
origin. The positive directions of these axes are usually taken
as indicated by the arrows in Fig. 256.
>x
Fig. 266
Fig. 257
Let P be any point in space, and let us consider the seg
ment OP. The numbers representing the projections of
OP on the three axes we call the coordinates of P and
denote them by x, y, and z. In Fig. 257, x = OA, y = OB,
Z=:00.
Conversely, any three real numbers x, y, z may be con
sidered as the coordinates of a point P. Why? If i^ is the
foot of the perpendicular dropped from P on the xyigleme, and
A is the foot of the perpendicular dropped from Mon the a>axis,
the coordinates of P are x — OA, y = AM, z = MP.
The eight portions of space separated by the coordinate
planes are called octants. From the preceding definitions it
496 MATHEMATICAL ANALYSIS [XXI, § 325
follows that the signs of the coordinates of a point P in any
octant are as follows :
(a) X is positive or negative according as P lies to the right
or left of the 2/2!plane ;
(6) y is positive or negative according as P lies in front or
back of the a;2;plane ;
(c) z is positive or negative according as P lies above or
below the ajyplane.
EXERCISES
1. What are the coordinates of the origin ?
2. What is the z coordinate of any point in the a;yplane ?
3. What are the x and y coordinates of any point on the 2:axis ?
4. What is the locus of points for which x = ? for which y = ?
for which ^ = ?
5. What is the locus of points for which x = and ?/ = ?
6. What is the locus of points for which y =0 and 2 = 0?
7. What is the locus of points for which s = and x = ?
8. What is the locus of points for which a; = 2 and y = 2 ?
9. If P(x, ?/, z) is any point in space, find
(a) its distance from the xyplane ; (d) its distance from the xaxis ;
(h) its distance from the ?/2;plane ; (e) its distance from the yaxis ;
(c) its distance from the x^plane ; (/) its distance from the ^axis.
10. Describe the positions of each of the following points : (2, — 8, 3) ;
(2, 3, 5); (3, 3, 3); (4, 7, 9).
11. Plot the following points : (2, 1, 3) ; (4, ~ 1,  2) ; (0, 0,  3) ;
(3,1,1); (1,1,1); (1,0,1); (1,2,1); (1,1,0);
(4,  1,  1).
12. Find the distance from the origin to the point P (a;, y, z).
13. A point P moves so that its distance from the origin is always
equal to 4. Find the equation of the locus of P.
14. Show that the points (a;, j/, z^ and (— x, y, z) are symmetric with
respect to the j/splane.
15. A rectangular parallelepiped has three of its faces in the coordi
nate planes. Find the coordinates of its vertices, assuming that the di
mensions of the parallelepiped are a, 6, c.
XXI, § 326]
LINEAR FUNCTIONS
497
Fig. 258
326. Directed Segments. We shall define the angle be
tween two directed lines I and m which do not meet, to be the
angle between two similarly directed
lines I' and m' which do meet (Fig. 258).
Theorem I.
If AB is a directed seg
ment on a line I, which makes an angle 6
with the directed line V, then
(1) Proiv AB==AB cos Q.
Proof : Through A' (Fig. 259) draw ?i parallel to I and let
Bi be the projection of B on l^. Then, by definition, the angle
between I and V is the same as the angle
between l^ and V. It follows from § 135
*^^* A^B'==A'B, cose,
or A'B' = AB cos By
since A'Bi = AB.
Theorem II. The projection on a directed line s of a broken
line made up of the segments A1A2, AzA^, ^3^4? •••, A^^iA^, is
equal to the projection on s of the segment AiA^.
The proof of this theorem is left as an exercise.
See § 136.
Corollary. If Pi (%, yi, z^) and P^ {X2, y%y Zz) are any two
points, then , ^ .
'X2  xi = Proja, P1P2,
(2) U2  yi = Projy P1P2,
^2  zi = Proj« P1P2.
EXERCISES
1. Find the projections upon the coordinate axes of the sides of the
polygon ABCDEF whose vertices are A (0, 0, 0), ^ (1,  6, 4), C ( 2,
4,  1), D (3,  1, 2), E{2, 1, 4), F (1, 1, 1).
2. The projections of the segment MP upon the coordinate axes are
4, 3,  1 respectively. If Jlf is (2,  1, 3), find the coordinates of P.
2k
498
MATHEMATICAL ANALYSIS [XXI, § 327
327. Direction Cosines of a Line. Let I be any directed
line and V a line through the origin having the same direction.
If V makes angles a, y8, and y with the x, y, and z axes respec
tively, then, by definition, I makes the same angles with these
axes. These angles are known as the direction angles of the
line Z, while their cosines are called the direction cosines of I.
Keversing the direction of a line changes the signs of the direc
tion cosines of the line. For reversing the direction of a line
changes a, /?, y into rr — a, ir — (3, tt — y, respectively ; and by
§122 cos(7r^) = cosa
Theorem. The sum of the squares of the direction cosines of
a line is equal to unity.
)»X
Fig. 260
Proof. Let P(x, y, z) be any point on V (Fig. 260). Then,
we have
{x = OP cos a,
(3) 2/ = OP cos ^,
[2; = OF cos y.
Therefore,
x^jy'' + z^= OP^[cos2 a + cos^ jS + cos^ y].
Since ips f 2/2 f gz = Op\* it follows that
(4) cos2 a +C0S2 p 4 cos2 7 = 1.
* or' = x2 + y2 and OP^ = 22 I oS^^ = ^2 I xa + ya.
± \/P + m2
+
n2
m
± VZ2 + 7^2
+
n^'
n
,
XXI, § 328] LINEAR FUNCTIONS 499
Any three numbers /, m, n, (not all zero) are proportional to
the direction cosines of some line ; for, P{1, m, n) is a point
and the direction cosines of OP are
I
cos a =
(5) cos (3
cos y =
±VZ* + m2+n2
The direction cosines of OP are evidently proportional to Z,
m, n, and they may be found by dividing I, m, and n, respec
tively, by ± V/2 + m^ 4 7i\
328. The Distance between Pi(xi, i^i, zj and ^2(^2, i/2, ^2).
Let the direction angles of the segment P1P2 be a, (3, y. Pro
jecting P1P2 upon the axes, we have, from the corollary of § 326,
P1P2 cos a = .T2 — Xi, P1P2 cos /? == 2/2 — 2/ij A A cos y = 22 — 2!l.
Squaring and adding we have, by the theorem of § 327,
i\P2"=(x2x,y^(y2yiyh(z2z,y.
Therefore,
(6) P,P2 = V(X2  X,y + (1/2  Vlf h (^2  ^l)^.
EXERCISES
1. Find the length and the direction cosines of the segment P1P2, when
(a) Pi is (2, 3, 4) and P2 is ( 1, 0, 5) ;
(&) Pi is ( 1, 2,  7), and P2 is (4, 1, 4) ;
(c) Pi is (4, 7, 1), and P2 is (1,  2,  7).
2. Prove that the triangle whose vertices are A{m^ w, p), 5(n, p, m),
C ( jo, m, n) is equilateral.
3. Find the direction cosines of a line which are proportional to 4, 7, 1.
4. Find the length of a linesegment whose projections on the co
ordinate axes are 4, 7, 2.
500 MATHEMATICAL ANALYSIS [XXI, § 329
329. The Angle between Two Directed Lines. If aj, p^, y^
and otg) ft, 72 ^^^ the direction angles of two directed lines
li and ^2 the angle 6 between them may be determined as
follows.
Draw the lines I'l and I'z through the origin, parallel to the
given lines (Fig. 261). Then the angle between l\ and V2 is 0.
^x
If P(x, y, z) is any point on l\, then, by Theorem II of § 326,
we have ^^^.^^^ ^^ ^ p^^.^^^ OMNF,
^'^' OP cos e = OM cos a2 + MN cos ft + iVP cos y^.
But,
OiJf = OP cos «!, il[fiV = OP cos ft, NP = OP cos yj.
Therefore,
(7) cos e = COS tti cos a2 H cos Pi cos P2 + cos 71 cos 72
We shall assume that 6 is the smallest positive angle satis
fying equation (7).
330. Parallel and Perpendicular Lines. If two lines are
parallel and extend in the same direction, they are parallel to
and agree in .direction with the same line through the origin.
Therefore, if ai, ^1, yi and ag) ft? 72 a^e the direction angles
of the two lines, a^ = a^, /3i = ft, 71 = 72 ; and we may write
(8) cos ai = cos a2, cos pi = cos p2, cos 71 = cos 72
Conversely, if relations (8) are satisfied, the given lines are
parallel and extend in the same direction. Why ?
XXI, § 331] LINEAR FUNCTIONS 501
If the two lines are parallel but extend in opposite directions,
we have ai = ir — 03, ft^ tt — ft? yi = 'r — 72? and therefore,
(9) cosai=cosa2, cos Pi=cos pg, cos 71 = cos 72
Conversely, if relations (9) are satisfied, the given lines are
parallel and extend in opposite directions. Why ?
If the two lines are perpendicular^ it follows from formula
(7) that,
(10) cos tti cos ttg + COS pi COS p2 + COS 7i cos 72 = 0.
Conversely, if (10) is true, the lines will be perpendicular.
If /, m, n and l\ m', n' are proportional to the direction
cosines of two lines, the lines will be perpendicular if, and
only if,
(11) W + mm' + nn' = 0.
They will be parallel if, and only if, the numbers I, m, n are
proportional to V, m', n'. If any of the numbers I, m, n are
zero, the corresponding numbers of the set Z', m', n' must, of
course, also be zero.
331. Point of Division. Let Pi(xi, 2/i> ^i), Afe Vii ^2) be
two given points and P{x, y, z) any point on the segment P1P2
P P
such that ^^ = X. If a, jS, y are the direction angles of this
PPi
segment, it follows from § 326 that
PiP cos a = x — aji, PP2 cos a= X2 — x.
Therefore,
PiP cos a __ X — Xi __,
PP2 cos a X2^ X
or
Xi +Xx2
(12) x =
1hX
502 MATHEMATICAL ANALYSIS [XXI, § 331
Similarly, we have
It should be noticed that A is positive if P lies within the
segment PiPo, and negative if it lies without. By varying A,
the coordinates of any point (=^ P2) on the line P1P2 may be
obtained.
For the midpoint of P1P2 we have A = 1 and, hence, the
coordinates of the midpoint of P1P2 are
(14) ^^ a;if a;2 ^ y=yi±Jh^ ^^ ^1 + 22 ^
i^ J 2
EXERCISES
1. Find the cosine of the angle between the two lines whose direction
cosines are proportional to 2, 3, 1 and — 1, 4, 6.
2. Find the coordinates of the points of trisection of the segment
Pi(4, 1,3), P2(4, 7,3).
3. Prove that the medians of the triangle whose vertices are (1, 2, 3),
(3, 2, 1), (2, 1, 3) meet in a point.
4. Show that the following points are the vertices of a right triangle :
(1,0,6), (7,3,4), (4, 5, 2).
6. If two of the direction angles of a line are 45° and 60°, find the
third direction angle.
6. Prove that the values a = 30°, /3 = 30° are impossible.
7. The direction cosines of a line are m, 2 m, 3 m. Find wi.
8. Show that {x—\Y+{y\ 2)2 f. (5; _ 3)2 = 9 is the equation of a
sphere whose center is at (1, — 2, 3) and whose radius is 3.
9. Express by an equation the fact that the point (x, y, z) is equi
distant from (2, 1, 3) and (— 1, 4, 3).
10. Show that the points (3, 7, 2), (4, 3, 1), (1, 6, 3), (2, 2, 2) are
the vertices of a parallelogram.
11. Prove by two methods that the points (3, 6, 4), (4, 13, 3),
(2, — 1, 6) are collinear.
12. Show that the points (4, 3,4), ( 2, 3, 2), ( 2, 9,  4) are
the vertices of an equilateral triangle.
XXI, § 332] LINEAR FUNCTIONS 503
13. Find the coordinates of the point which divides the segment Pi Pa
in the ratio X, given
(a) Pi(2, 6, 8), P2(l, 3, 6),X = 3;
(6) Pi( 2,  5, 8), P2(8, 0,  2), X =  2 ;
(c) Pi(3,  7,  9), P2(2,  2,  1), X = 1.
14. Prove that the medians of the triangle Pi(a;i, yi, ^i), P2(X2, y2, ^2)1
Ps(x3, ys» Z3) meet in the point
( xi + ^2 + a;3 yi + y2 + Vs gi + g2 + ss \
V 3 ' 3 ' 3 /
15. Prove that the lines joining the midpoints of the opposite edges
of a tetrahedron pass through a common point and are bisected by that
point.
16. Are the following points collinear : (2, 1, 3), (—2, —5, 3),
(1, 5, 7) ?
17. Find the direction cosines of the line that is equally inclined to the
three axes.
18. Prove that the lines joining successively the middle points of the
sides of any quadrilateral form a parallelogram.
19. Find the projection of the segment Pi(l, 2, 3), P2(2, 1, 3) upon
the line that passes through the points P3(— 3, 5, — 5) , P4(8, — 9, 12).
332. Locus of an Equation. We saw that in the plane the
locus of the equation f(x, y) =0 represents, in general, a curve.
In an analogous way the equation f(x, y, z) = 0, in general, re
presents a surface. For, if we solve for z, we have z = F(x, y)
and from this equation, we see that we can find, corresponding
to every point (x, y) in the xy^lame, one or more values of z
(real or imaginary). The locus of the real points (x, y, z) is,
in general, a surface, but may be a curve or a point. If there
are no real values for a;, ?/, z which satisfy the equation
f{x, y, z) — 0, we say that the equation has no locus.
The locus of points satisfying the two conditions /(a;, y, 2!)=0
and F{x^ y^z)—^ is, in general, a curve in space, which is the
intersection of the two surfaces represented by these equations.
504 MATHEMATICAL ANALYSIS [XXI, § 333
333. The Plane. A plane is defined as a surface such that
every point collinear with two points of the surface is itself a
point of the surface.
We shall prove the following propositions :
(a) Every equation of the first degree in x, y, and z represents
a plane.
ih) Every plane is represented by an equation of the first de
gree in X, y, z.
To prove (a), let Pj (xi, yi, z^), P^ixz, 2/2? ^2) be any two points
on the surface whose equation is Ax { By + Cz \ D = 0.
Then we have
(15) Ax^ + By,+ Cz^ + D^(),
(16) Ax^ 4 By2 h Cz^ + 2) = 0.
Now let Pi{x^, 2/3J 2:3) be any point on the line PiP^ Then
(if P3 ^fc Pa)) there exists a value of A.(=5fc — 1) such that
"^"TTT' ^^TTT' '^"TTX" ^^^^^^
We wish to show that the coordinates of this point also satisfy
the equation Ax \ By \ Cz + D == 0. By substitution in this
equation we have
(17) .^{Ax,\ By^\Cz,\D)^,:^^{Ax^^By^^Cz2+D) = 0.
Relation (17) is true, since it follows from (15) and (16) that
each parenthesis vanishes separately. Therefore the surface
defined by the equation Ax \ By ■\ Cz \ D = satisfies the
definition of a plane.
To prove the statement (6), let tt be any plane, and let OH
be the perpendicular from which meets ir in Pj (Fig.
262). The positive direction oi OR will be taken from to
the plane. The direction angles of OR will be called a, )3, y
XXI, § 333J
LINEAR FUNCTIONS
505
and the length OPi will be denoted by p.* Now if P(Xj y, z)
is any point in the plane, we have, by § 326,
>3:
Fig. 262
(18) Projo50P= Tvo]onOM+ ViojonMN^ ProJo^iVP.
Hence the equation
(19) X cos a 4 y cos p + z cos y = p
is the equation of the plane. Why ? It is seen to be an equa
tion of the first degree in x, y, z. This form of the equation
is called the normal form.
It follows from the above that, if Ax ^^ By \ Cz \ D = i^
the equation of a plane, the direction cosines of a line perpen
dicular to the plane are proportional to A, B, C.
It is left as an exercise to prove that to reduce Ax \ By \
C^ f D = to the normal form we must divide each term by
± V^^ + B^ + C% the sign of the radical being chosen opposite
to that oi D if D =^ 0, the same as that of C ii D = 0, the
same as that of B if C=D = or the same as that of A if
B=C=D = 0*
* If the plane passes throilgh the origin we shall suppose OR is directed
upward, and hence cos 7 >0 since 7 <7r/2. If the plane passes through the
zaxis, then OR lies in the ccyplane and cos7 = 0; in this case we shall sup
pose OR so directed that /3<7r/2 and hence cos/3>0. Finally if the plane
coincides with the yzplane, the positive direction on OR shall be taken as
that on OX
^2
±V^o^
B,
■vc^'
±V^2^
+ 02^
506 MATHEMATICAL ANALYSIS [XXI, § 334
334. The Angle Between Two Planes. The angle between
two planes is defined to be the angle between two normals {i.e.
perpendiculars) to the planes. Let AiX \ B^y + Cis; + Di =
and A^x + Bay \ C2Z { D2 = ^ be the equations of the two
planes. The direction cosines of their normals are then (§ 333),
cos «! = ^ , cos 02 =
cos Bi = ^ cos ^2 =
cos yi = ^ ; cos y2 =
If ^ is the angle between these normals, then, from § 329,
(20) cos ^ = ± A,A 2±BA + C,C2
V^l^ + B{^ + Ci2 V^z^ + ^2^ + C22
If the planes are perpendicular, cos ^ = 0, and we have
(21) A^A^ + B1B2 4 Ci(72 = 0.
If the planes are parallel their normals are parallel. Hence,
by § 330, their equations in normal form are
X cos a\y cosySjg cos y—p, x cos «'+ ?/ cosy8'+2;cos y'=p',
where either cos a = cos a', cos ft = cos ft', cos y = cos y', or
cos a = — cos a', cos y8 = — cos ft', cos y = — cos y'. Therefore,
if the two equations be written in the form
A^x + Biy + Ciz + Di = 0, AzX + ^22/ + C^z + A = 0, .
the planes will be parallel if and only if
(22) A2 = kA,, B2 = kB,, Cz^kCi. (k^O)
The equation of any plane parallel to Ax \ By + Cz \ D =
can therefore be written in the form Ax ^ By + Cz + D' = 0.
XXI, § 334] LINEAR FUNCTIONS 507
EXERCISES
1. Sketch the planes whose equations are (a) a; =2, (6) y = 4,
{c) z=6,(id) 2x + y = l, (e)yz = 0.
2. How many arbitrary constants are there in the equation of the
plane Ax+By+Cz + D = 0?
3. What is the general equation of a plane that passes through the
origin ?
4. What is the equation of the x^zplane ? y«plane ? a;2!plane ?
5. What are the intercepts on the axes of the planes whose equations
are
(o) 2x3y\z = 12', (6) xy+z=8; (c) x + y = 0; (d) 6x7=0?
6. Give three numbers proportional to the direction cosines of the
normal to the plane x\2y — z = 9. What are the direction cosines ?
7. What is the normal equation of the plane x — y + z = 9?
8. What is the equation of the system of planes parallel to
2x — y + 2! = 1?
9. What is the equation of the plane that passes through the origin
and is parallel to 2x — Sy + 7z = b?
10. Show that the planes 2x + 4ty — z = 2 and 4a;— yf40 = 7 are
perpendicular.
11. What is the equation of the plane parallel to2x + 2y\z=:9 and
6 units farther from the origin ? 2 units nearer ?
12. What is the distance between the parallel planes 2xi2y + z = 9
and 2 X + 2 ?/ + « = 15 ?
13. Find the equation of the plane passing through the points
(a) (1,2,1), (1,1,0), (0,0,1);
(&) (2,1,3), (1,1,2), (1,1,4);
(c) (2,2,2), (1,1, 2), (1,1,0);
(d) (1,1,1), (1,1,2), (2,2,2).
[ Hint : Use the equation Ax\By+Cz + I)=0 and divide by any coeffi
cient that is not zero.]
14. If D 9^ show that the equation Ax \ By + Cz ■} D = can be
written in the form x/a{y/b + z/c = l where a, b, c are the inter
cepts made by the plane on the x, y, z axes respectively.
15. Show that the four points (0, 0, 3), (4, —3, —9), (2, 1, 2),
(4, 3, 3) are coplanar, i.e. they lie in the same plane.
508 MATHEMATICAL ANALYSIS [XXI, § 334
16. Find the equation of the plane that passes through the point P
and is parallel to the plane a, when
(a) Pis (2, 1, 8) and ais2£c + 3y 50 = 5 ;
(6) P is (1, 0, 0) and a is 2« + 2/ + = 1 ;
(c) Pis (2,  1,6) and a is 3a; 5 2/ 2« = 3.
17. Find the equation of the plane passing through the point P and
perpendicular to the planes a and /3 when
(a) P is (1, 1, 1), a is 2 x — 2/ — = 4, and ^isx^y + z = l;
(b) Pis (1, 2, 1), a is a; + 2/ 3« = 3, and /3 is .Sx 5y + 22 = 1 ;
(c) Pis (0, 3, 4), ais2a: + 4y + = 7, and /3is2x0 + 32! = 2.
18. Find the equation of the plane passing through the points Pi, Pa
and perpendicular to the plane a, when
(a) Pi is (1, 1, 1), P2is (1,2, 1), and a is 2x 3y  ;s = 2;
(b) Pi is (0, 0, 1), P2 is (2, 1, 3), and aisx + y — 6;s = 0;
(c) Pi is (2, 1,  3), P2 is (0, 4, 2), and «is4x — y — ^ = 2.
19. Prove that the distance from the plane Ax + By + Cfe + D = to
the point (XI, 2/1, 01) is + Ax, + By,+ Cz, + D^
VA^ + P2 + C=2
10. Find the distance from the plane a to the point P when
(a) P is (2, 1, 4) and ais2x — iy }■ z = 2 ;
(6) Pis (2,3,  1) and a is 2x + 2/ + 2602 = 0;
(c) P is (0, 0, 3) and ais3x22/50 = l.
21. Prove that the equation of the plane which passes through the
point (xi, 2/1, «i) and is parallel to the plane Ax + By \ Cz {■ D = is
A(x  xi) + B{3j  2/1) + C{z  zi) = 0.
22. Prove that the equation of a plane which passes through the
point (xi, 2/1, zi) and is perpendicular to the plane Ax+By+Cz+D =
is (P01— C2/i)x+ (Cxi— ^01)2/+ (^2/1 — Bxi)z = 0.
23. Find the cosines of the angles between the following pairs of planes
(a) 2x32/ + = l, 2x + z=z0;
(6) xy0 = 2, 2^40 = 8;
(c)x + = 3, 4x + 2/+30 = 5.
24. Find the equation of the plane that passes through Pi, P2 and
makes an angle with the plane a, where
(a) Pi is (0, 1, 0), P2 is (0, 0, 1), a is 2/+ 07 = 0, and d is 120°;
(6) Pi is (1, 0, 1), P2 is (0, 1, 2), a is X + 2 2/ + 20 = 2, and 6 is 60°.
XXI, § 335] LINEAR FUNCTIONS 509
25. Find the equation of the locus of a point which moves so that its
distance from the x?/plane is twice its distance from the a^axis.
26. Find the equation of the locus of a point whose distance from the
plane x + 2y — 5 = 0is twice its distance from the za,xis.
27. A point moves so that its distance from the origin is equal to its
distance from the ^rxplane. Find the equation of its locus.
335. Simultaneous Linear Equations. In § 70 we saw that
three simultaneous linear equations in three unknowns have
in general a single solution. We shall now show that three
such simultaneous equations have either, (a) a single solu
tion, or (6) an infinite number of solutions, or (c) no solution.
We shall prove this statement geometrically. Each equa
tion represents a plane ; the three planes may assume the fol
lowing relative positions.
Case I. No two of the planes are parallel or coincident.
(a) The three planes may intersect in a single point ; then
there is a single solution of the three simultaneous equations.
(&) The three planes may intersect in a line ; then there is
an infinite number of solutions.
(c) The three planes may intersect so that the three lines of
intersection are parallel ; then there is no solution.
Case II. Two of the planes are parallel but not coincident.
In this case the three planes can have no point in common
and the equations have no solution.
Case III. Two of the planes are coincident.
(a) The third plane may be parallel to the coincident planes,
in which case there is no solution.
(6) The third plane may intersect the coincident planes, in
which case there is an infinite number of solutions.
(c) The third plane may coincide with the coincident planes,
in which case there is an infinite number of solutions.
510 MATHEMATICAL ANALYSIS [XXI, § 336
336. Pencil of Planes. All the planes that pass through a
given line are said to form a pencil of planes. If
A,x + B,y + Ciz H A = 0,
(23)
^ [A2X + B^y + C^z f A = 0,
are the equations of any two planes passing through the given
line, then the equation of any other plane of the pencil can be
written in the form
(24) A^x + B^y j^C,z+D,+\ {A^x h B^ + C<,z + A) = 0,
where A is a constant whose value determines the particular
plane of the pencil. (See § 68.)
337. Bundle of Planes. All the planes that pass through a
common point are said to form a bundle of planes, and this
common point is called the center of the bundle. If
' A^x + Biy+C^z + D^ = 0,
(25) A2X + A2/ + C2Z + A = 0,
A^x + B,y^C^z+ A0,
are the equations of any three planes passing through the
center and not belonging to the same pencil, then the equation
of any other plane of the bundle is
(26) {A^x + Biy + C^z + D^) + X^A^x + B^y + C^z + A)
+ X2{Ax 4 B,y + C> + A) = 0,
where Ai, A2 are constants whose values determine the position
of the particular plane of the bundle. Why ?
EXERCISES
1. Find the equation of the plane that passes through the Intersection
of the planes a and /3 and the point P, when
(a) ais2x + 3y2; = l, /3isa; + 2/2« = 2, and P is (1, 0, 2) j
(6) aisx + y + 2^ = 0, /3is4a;2?/2; = l, and Pis (2, 1,1);
(c) a is 3 a:  2 y  2r = 2, /3 is X  2/ + « = 3, and P is (1, 0, 1).
XXI, § 338] LINEAR FUNCTIONS 511
2. Show that the planes whose equations are 3a;5y + 2 = 0, 6x +
y = 2 2 + 13, lly — 2z = 17, belong to the same pencil.
3. What is the equation of the plane of the pencil whose axis is
2x — y + 52! + 2 = 0, 4x — Sy\z = l, which is perpendicular to the
plane x = 0? y = 0? z = 0?
4. Find the equation of the plane that passes through the intersection
of the planes 2x + y — z j 1, Sx— y — z = 2 and is perpendicular to the
plane x ^ y — z = 1.
5. Find the equation of the plane that passes through the point of in
tersection of the planes a, /3, y and the points Pi, P2, when
(a) ais2x + y = 1, fi is x — z = 1, yis2x — y + 2z = 3,
Pi is (1, 0, 1), and P2 is (2, 1, 1) ;
(b) aisSx y z = S, ^isx — y\2z = l, VisSx — 2y + 2! = 3,
Pi is (2, 1, 3), and P2 is (0, 8, 0).
338. Equations of a Straight Line, (a) The two simul
taneous equations
(27)
^ ^ ^ A2X + B^yh C2Z } A = 0,
represent a line, the intersection of the two planes, provided
the two planes are not parallel.
(6) A given point and a given direction determine a line.
Let the given point be Pi{xi, y^, z^ and a, /8, y the given direc
tion angles. If P (x, y, z) is any other point on the line at a
distance d from Pj, then by § 326, dcos a=x—Xijd cos /8=2/— 2/ij
d cos y = z — Zi. Hence we may write
(28) xx^ ^ yyi ^ z z^ ^
cos a cos p cos y *
which are the equations of the required straight line. These
equations are known as the symmetric equations of a straight
line. In these equations cos a, cos /8, cos y can evidently be
replaced by ar.y three numbers proportional to them.
512 MATHEMATICAL ANALYSIS [XXI, § 338
(c) Two distinct points Pi(xi,yi, Zi), P^ixz, 2/2? %) determine a
line. Any line through, the point Pj is of the form
cos a cos ^ cos y
Now the direction cosines of PiA are proportional to a^ — ajj,
?/2 — J/i> 2^2 — 2:1. (§ 328.) Therefore the equations of the line
through the points Pi, P.^ are
(29) xXi ^ y Vi ^ zZi
X2 Xi Vi Vi ^2  2i
We should note that in every case two equations are necessary
to represent a line.
Example 1. Reduce to the symmetric form the equations of the straight
line, 2a: + y0 = 3, x — y+2z=:1. Eliminating y between the two
equations we have 3 a; + = 10. Similarly, eliminating z we have
6 aj f y = 13. Solving these two equations for x and equating the values
found, we have
a;_ y — 13 _ g10
1~ 6 ~ 3 *
The line is seen to pass through the point (0, 13, 10) and to have direction
cosines proportional to 1,  5, — 3.
Example 2. Find the equations of the line that passes through the
point (4, —1, 3) and is perpendicular to the plane 2aj — 3y + 40=7.
The required line is parallel to any line perpendicular to the plane and hence
its direction cosines are proportional to 2, —3, 4 (§333). Therefore,
the equation of the required line is
a;~4 _ y + l _ g3
2 3 ~ 4 *
EXERCISES
1. Write the equations of the line that passes through the point P
and whose direction cosines are proportional to a, 6, c, where
(a) Pis (1, 2, 1) and a = 2, 6 =  7, c = 2;
(6) P is (3, 0,1) and a 2, 6 = 8, c = 9;
(c) P is (3,  2, 6) and a = 2, 6 =  9, c = 3.
XXI, § 338] LINEAR FUNCTIONS 513
2. Find the equations of the lines passing through the following pairs
of points :
(a) (2, 1, 4), (12, 2, 8) ; (c) (4, 3, 8), (8,  2, 1) ;
(6) (3,  6,  3), ( 3, 5, 7) ; (d) (5, 2, 1), (4, 7, 9).
3. Write in symmetric form the equations of the lines
(a) 2x7j + Sz = S, Sx + by ^z=:9;
(b) Sxyz = 8, 4:X + Qy5z = S;
(c) 5x\Sy + z=S, 2x — y + z = 'l.
4. Find the equations of the line that passes through the point P and
is perpendicular to the plane a, when
(a) Pis (2, 1, 7) and ais3ic?/ + 40 = 9;
(6) P is (4, 2,  2), and a is 2 a;  6 2/ + 3 = 3 ;
(c) Pis (— 1, 6, 3), and a is 3a; + 4?/ — ;s = 5.
6. Find the equations of the line that passes through the point
(2,— 1, 4) and is parallel to the line
x_3_ y7 _ g7
4 ~ 2 3 '
6. Find in symmetric form the equations of the line that passes
through the point (2, — 1, 4) and is parallel to the line 2x + y — z = 6,
X — y + Zz = 4:.
7. Find the equation of the plane that passes through the point P and
is perpendicular to the line Z, when
(a) Pis(2,5, 1), andns^ = ^li = ^;
o o o
(6) Pis ( 1, 4,7), and I is =^ = 1^ = ^±f.
8. Find the equation of the plane that passes through P(l, 5, 2) and
is perpendicular to the line 3xy+2 = 8,x — y + 22! = 6.
9 If tf is the angle between the two lines ^ ~ ^^ = V^rJll  lull^
«2 62 C2
10. Prove that the lines  = ^ = ^ , ~ = ^ =  are perpendicular
to each other. 624463
2h
CHAPTER XXII
QUADRATIC FUNCTIONS. QUADRIC SURFACES
339. The Sphere. If a point P{x, y, z) moves so as to be
always at a constant distance r (r > 0) from a fixed point {h, k, I),
the locus of P is called a sphere. The equation of this locus is
(1) (xhy+(yky{.(ziy = r^.
If this equation is expanded, it has the form
(2) x^\y^\z^\Ax{By + Cz^D = 0,
where A, B, C, D are constants depending upon the coordinates
of the center and the length of the radius.
Conversely, an equation of the form (2), in general, repre
sents a sphere, for it can be written in the form
<" ('+f)'('f)'H'+f)'=t'f?^
which is a sphere if
4 4 4
The center of the sphere is at the point (— A/2, — B/2, — 0/2),
and the radius is
V^V4 + S2/4 + 074  D.
If the righthand member of (3) is zero, the locus is the
single point (— Ji/2, —B/2, — (7/2). If the righthand member
of (3) is less than zero, the equation has no locus. See § 206.
614
XXII, § 340] QUADRIC SURFACES 515
EXERCISES
1. Find the equation of the sphere whose center is at P and whose
radius is r, when
(a) Pis (2, 1, 9),andr = 6;
(6) Pis(l, 8,0), andr = 2;
(c) P is (4, 9,2), and r = 7.
2. Find the equations of the eight spheres tangent to the three co
ordinate planes and having a radius of 4,
3. Find the equation of the sphere which has the line joining P(2, 6, 8)
and ^(4, 6, 6) as a diameter.
4. Discuss the locus of each of the following equations,
(a) x2 + ?/2 + ^2 _ 2 X  2 y  2 2 = 6 .
(ft) a;2 + y2 _}. 2.2 _. 4 a; ___ 4 y _ 6 2 + 25 = 0.
(c) a;2 + 2/2 + ^2 _ 2 X  6y + 8 = 5.
{d) x2 + 2/2 f ;22 _ 2 2  4 y + 5 = 0.
5. Find the locus of points the ratio of whose distances from (0, 1, 0)
and (1, 2, 3) is 5.
6. Show that the equation of the tangent plane to the sphere
x'i + y^^Z^ =^2
at the point (xi, j/i, zi) is
xxi f yyi + zzi = r2.
[Hint : The tangent plane is perpendicular to the radius.]
7. Find the equation of the sphere passing through the following
four points.
(a) (1,2,3), (3,1,0), (2,1,0), (3,4, 1).
(6) (2, 1, 0), ( 1,  1, 0), (3, 0, 2), (0, 0,0).
[Hint : Use the equation x^ h y^ ^ z'^ + Ax ] By \ Cz + D = and
determine the values of A, J5, O, Z>.]
340. Cylinders. The surface generated by a straight line
which moves parallel to a given line and always intersects a
given fixed curve, is called a cylindrical surface or a cylinder.
The generating line in any of its positions is called an element
of the cylinder.
Any algebraic equation in two cartesian coordinates repre
sents in space a cylinder whose elements are parallel to the
axis of the third variable.
516 MATHEMATICAL ANALYSIS [XXII, § 340
For example, the equation
a;2 + y2 =4
represents in the ccyplane a circle (Fig. 263). But, the equation is satis
fied by the coordinates of any point P which lies on a line parallel to the
«r
2;axi8 and passes through a point Q on the circle. Moreover, if QP moves
parallel to the 2axis and continues to cut the circle the coordinates of P
still satisfy the equation x^ + y2 _ 4, The cylinder traced by the line
QP is the locus of the equation x^ + 2^2 _ 4,
It is clear that if a cylinder has its axis parallel to a coordinate axis, a
section made by a plane perpendicular to that axis is a curve parallel
and equal to the directing curve on the coordinate plane. Thus the
section cut by the plane z = S from the hyperbolic cylinder whose equa
tion is
x^y^ = 4,
is a hyperbola equal and parallel to the hyperbola in the xyplane whose
equation is x^ — y^ = 4.
341. The Projecting Cylinders of a Curve. A cylinder
whose elements are parallel to one of the coordinate axes and
always intersect a fixed curve in space, is called a projecting
cylinder of the curve. The equations of the projecting cylin
ders may be found by eliminating in turn each of the variables
X, y, z, from the equations of the curve. Why ? The curve
may often be constructed conveniently by means of two dis
tinct projecting cylinders.
XXII, §342] QUADRIC SURFACES 517
EXERCISES
1. Describe the locus of each of the following equations,
(a) aj = 2. (A) yz  6.
(6) 2x2 + y2 = 8. (4) ^_^ = o.
^^^ ^^ ^' (m) y2==3c2.
(fir) 2/2  ig2 = 1.
2. Prove that x^ + 2xy +y^ = 1 — z^ is the equation of a cylinder,
the direction cosines of any element being proportional to (1, 1, 0).
3. Find the equations of the projecting cylinders of each of the following
curves. Construct the curves as the intersection of two of these cylinders.
(a) x2 + y2 + 2,2 = 4^ a;2 + y2 _ 5,2 = 0.
(6) a; = 1, x2 + y2 4. ;5,2 = 4.
{c) x^  y^ = ^ Zj x^ + y^ = z.
{d) y'2 = x + z,z = x + yK
(e) 02 = xy, x2 = yz.
342. Symmetry, Intercepts, Traces, Sections. If a given
equation is unaffected by replacing x hj — x throughout, the
locus is symmetric with respect to the yz^lane.
If a given equation is unaffected by replacing y by — y, the
locus is symmetric with respect to the a;2;plane.
If a given equation is unaffected by replacing zhj — z, the
locus is symmetric with respect to the xyi^\sine.
What would be a test for symmetry with respect to the a;axis ? the
t/axis ? the «axis ? the origin ?
The segments measured from the origin to where a surface
cuts the axes are called the intercepts of the surface on the
axes. To find the intercepts place two of the variables equal
to zero and solve the resulting equation for the third variable.
Why?
518
MATHEMATICAL ANALYSIS [XXII, § 342
The sections of a surface made by the coordinate planes are
called the traces of the surface (Fig. 264). To find the equa
tions of the traces put each variable in
turn in the given equation equal :to zero.
Why?
The equations f(x, y, z)—0 and x = k,
a constant, are together the equations of
the curve of intersection of the surface
and a plane parallel to the i/^^plane.
Similarly sections parallel to the xy and 2/zplanes may be
found. If A: = 0, the sections are the traces.
*x
343. The Ellipsoid. The surface represented by the equation
w
1' 4.1^' 4.^=1
ai b^ c2
is called a,ii ellipsoid. It is symmetric
with respect to the three coordinate
planes, the three axes, and the origin.
The intercepts on the x, y, saxes are
respectively ± a, ±b, ± c (Fig. 266).*
The traces on the three coordinate planes
are, respectively.
x^_^y
&2
1, . = 0; ^' + ?!=l, 2/
0;
Fig. 265
1, X = 0.
The sections of the ellipsoid by the plane a; = A; is an ellipse
whose equations are
(''f<'S)
= 1, xz=k.
bVl
* The figure exhibits only that part of the surface lying in one octant,—
that in which x, y, z are all positive.
XXII, § 343]
QUADRIC SURFACES
519
The semiaxes of this ellipse are 6 VI — Wja^, cVT^F/o^.
As I A; I increases from to a, the axes of this elliptical section
decrease. When  A;  = a the ellipse reduces to a point, and
when I A; I > a the sections are imaginary. The surface lies
therefore entirely between the planes x — a^ x = — a. Sim
ilarly it may be shown that the surface is also bounded by the
planes y = b, y — — b ; z = c, z = — r.
LI
'
n
^
Fig. 26()
A general idea of the appearance of an ellipsoid is given
by Fig. 266, which represents a plaster model of this sur
face.
Special Cases. In general the semiaxes a, b, c are unequal,
but it may happen that two or three of them are equal. If
the three are equal, i.e. a = b = c, the surface is a sphere. If
two are equal, for example, if & = c, the ellipsoid is called an
ellipsoid of revolution, for it can be generated by revolving the
ellipse x'^/a^ \ y^/b^ = 1, j; = about the avaxis.
520 MATHEMATICAL ANALYSIS [XXII, § 344
344. Surfaces of Revolution. The surface generated by
revolving a plane curve about a line in its plane is called a
surface of revolution. The equation of the surface is readily
found when the axis of revolution, i.e. the line about which
the curve is revolved, is one of the coordinate axes.
Let y=f{x) be the equation of the
plane curve in the a;?/plane and the
icaxis the axis of revolution. As the
curve 2/= /(a?) revolves about the a;axis,
any point P on this curve describes a
circle, whose center is on the xaxis and
whose radius is equal to f{x) (Fig. 267).
Therefore for any position of P (x, y, z) we have,
which is the equation of the required surface of revolution.
If the ellipse rK'^/a^ f y^/h^ = 1, ^ = is revolved about the xaxis, the
equation of the surface of revolution is
^^ + ^^ = ita^^^3, or 1 + 1^,4^=1.
EXERCISES
1. Sketch and discuss each of the following ellipsoida.
(a) 9 a:2 + 4 2/2 + 16 z^ = 144.
(6) 25 a;2 + y2 _i_ ;j;2 =: loo.
(C) x2 + 8y2 + 2 2;2=:16.
2. Show that the ellipsoid in Ex. 1 (6) is an ellipsoid of revolution.
3. Find the equations of the ellipsoids formed by revolving the follow
ing ellipses about the axes mentioned.
(a) 9 x2 + 4 y2 = 36, « = o, xaxis.
(6) 9 x2 + 4 y2 = 36, « = 0, 2/axis.
(c) 9 x2 + «2 = 9, ?/ = 0, ^axis.
(d) 25 2/2 + 4 22 = 100, X = 0, ?/axi8.
XXII, § 345]
QUADRIC SURFACES
521
4. When an ellipse is revolved about its major axis the ellipsoid gen
erated is called a prolate spheroid; when it is revolved about its minor
axis, an oblate spheroid. Which of the ellipsoids in Ex. 3 are oblate and
which are prolate ?
5. Describe the locus of each of the following equations.
(a) ^ + ^ + 5_ = o
^ ^ a^ b^ c''
^ ^ a^ b^ d^
345. The Hyperboloid of One Sheet. The surf ace. repre
sented by the equation
(6)
Qi "^ 62 ^2
is called a hyperboloid of one sheet
respect to each of the coordinate
planes, each of the coordinate
axes, and the origin. The inter
cepts on the X and ?/axes are
± a and ± b respectively, while
the surface does not meet the
zaxis (Fig. 268). The traces on
the coordinate planes are, respec
tively,
It is symmetric with
Fig. 268
 = 1, 2/=0.
a2 c2 ' ^
Of these, the trace on the xy^lsme is an ellipse, while the other
two are hyperbolas.
The section of the surface made by the plane z = k, is an
ellipse whose equations are
•t'?]''t'n
= 1, z = A;.
522
MATHEMATICAL ANALYSIS [XXII, § 345
This ellipse is real for all real values of k. The semiaxes are
the smallest when A; = and increase without limit as  A: 
increases.
The plane y = X,  A.  ^ 6, intersects the surface in the
hyperbola
= 1, 2/ = X.
•■['^] I's]
If I A I < 6 the transverse axis is parallel to the icaxis, while
if I A I > 6 it is parallel to the 2;axis.
Fig. 209
A good idea of the appearance of this surface is given by
Fig. 269, which represents a plaster model of a portion of the
surface.
If A, = 6, the section consists of the two straight lines
a c
a c
XXII, § 345] QUADRIC SURFACES 523
If A. = — 6, the section is the two lines
a c a c
These four straight lines lie entirely upon the surface.
Similar considerations apply to the sections made by planes
parallel to the yz^lsme.
. The form of one eighth of the surface is given in Fig. 268.
The broken lines in that figure indicate three sections by the
three planes
y = \, for I A, I < 6, =b, and > b.
Some of the straight lines on the surface are shown on the
model represented by Fig. 269.
If a = 6 the hyperboloid becomes a surface of revolution
obtained by revolving the hyperbola x^/a^ — z^/c^ = 1, y =zO
about its conjugate axis.
EXERCISES
1. Sketch and discuss each of the following surfaces.
(a) x2+4 2/2_g2=16. (ft) 9x^\y^z^=Se. (c) ix^+16y^z^=Qi.
2. Are any of the surfaces in Ex. 1 surfaces of revolution ?
3. Show that
(x2y (yl)2 (zSr ^^
9 4 1
is the equation of a hyperboloid of one sheet whose center is at the point
(2, 1, 3).
4. Show that ^t: + ?l = i and ^ + l^+£!=3l are equations of
a'^ b'^ c^ a'^ 52 ^2
hyperboloids of one sheet.
5. Find the equation of the hyperboloid of revolution formed by
revolving each of the following hyperbolas about the axis specified.
(a) 9 a;2  4 1/2 = 36, z = 0, transverse axis.
(6) 9 x2  4 2/2 = 36, ;j; _ 0, conjugate axis.
(c) 4 1/2 _ 2;2 _. 16, a; = 0, transverse axis.
(d) 4 1/2. _ z2 = 16, X = 0, conjugate axis.
524
MATHEMATICAL ANALYSIS [XXII, § 346
346. Hyperboloid of Two Sheets. The surface represented
by the equation
(6)
fl2 e,2
22
C2
==1
is called a hyperboloid of two sheets. It is symmetric with
^x
Fig. 270
respect to each of the coordinate planes, the coordinate axes
and the origin. The intercepts on the a?axis are ± a, while
Fig. 271
the surface does not meet the y or zaxis (Fig. 270). The
traces on the coordinate planes are, respectively.
a;2 y'
There is no trace on the 2/2?plane.
XXII, § 346] QUADRIC SURFACES 525
The plane a; = A; intersects the surface in the curve whose
equations, iik^± a, are
11'] {5]
1, x = k.
If I A: I > a this curve is an ellipse ; if \k\ = a it is a point.
If 1 A: I < a the equations have no locus. All sections parallel
to the xy and a;2;planes are hyperbolas.
A good idea of the appearance of this surface is given by Fig.
271, which represents a model of a portion of the surface.
If 6 = c the hyperboloid becomes a surface of revolution
formed by revolving the hyperbola
X
about its transverse axis.
EXERCISES
1. Construct and discuss each of the following surfaces,
(a) 4 a;2 ~ 9 ^2 _ 36 ^2 = 144.
(6) x2  y2 _ z2  1.
(c) 9x24«/202 = 36.
2. Are any»of the surfaces in Ex. 1 surfaces of revolution ?
3. Show that _ ± + ^  5 = 1 and _ ± _ ^^ + ^ = 1 are equations
a2 &2 c2 02 62 ^c2 ^
of hyperboloids of two sheets.
4. Find the equation of the hyperboloid of revolution formed by
revolving each of the following hyperbolas about the axis specified.
(a) ^  ^ = 1, = 0, conjugate axis.
4 9
(6) 4:y^  z^=4, x = 0, transverse axis.
(c) 2 x2 — 4 02 = 1^ y = 0, conjugate axis.
526
MATHEMATICAL ANALYSIS [XXII, § 347
347. The Elliptic Paraboloid. The surface represented by
the equation
is called an elliptic paraboloid. It is symmetric with respect
to the xz and 2/2jplanes, and the 2axis. The
intercepts on all three axes are zero. The trace
on the a;yplane is a point, namely the origin ;
the traces on the xz and yziplsmes are, respec
tively, the parabolas x^ = a^z, y = 0; y^ = b^z,
x = (Fig. 272).
Sections made by the planes z = k (k > 0)
are ellipses. Why ? Those made by the planes x = k and
y = kj respectively, are parabolas. Why ?
Fig. 272
FiQ. 273
Figure 273 represents a model of a portion of the surface.
If a = 6, the surface is a figure of revolution formed by re
volving the parabola x^ = a% y = 0, about the 2;axis.
XXII, § 348]
QUADRIG SURFACES
527
348. The Hyperbolic Paraboloid. The surface represented
by the equation
x2 y2
(8)
fe2
is called a hyperbolic paraboloid. (See Fig. 274.) It is sym
metric with respect to the xz and yziplsmes. All three inter
cepts are zero. The trace on the xyiplsme is the pair of lines
^ . ?/
0, ^=0;
the traces on the xz and t/^planes are, respectively, the
parabolas
x^ = a% y = 0; y^ =  b% a; = 0.
Fig. 274
Fig. 275
Sections parallel to the a;?/plane are hyperbolas, while those
parallel to the xz and 2/2!planes are parabolas. The form of
the surface is shown in Fig. ?75.
528 MATHEMATICAL ANALYSIS [XXII, § 348
EXERCISES
1. Sketch and discuss each of the following surfaces,
(a) a;2 + 4y2 = 36 z. (c) y'^  z^ = x.
(6) 2X2 + ;32 = 16y. (d) 2X2  02 = _ y.
2. Sketch the surface «22a; + t^4y = «~5.
349. The Cone. The surface represented by the equation
is called a cone. It is symmetric with respect to the three
coordinate planes, the three axes, and the origin. All three
intercepts are zero. The trace on the xy
plane is a point, namely the origin. The
traces on the x% and 2/2!planes are respec
tively the pairs of lines ca; ± az = 0, 2/ = ;
c?/ ± 6^ = 0, a; = (Fig. 276). Sections paral
lel to the a;?/plane are ellipses, while those
parallel to the xz and 2/2!planea are hyper
bolas. If any point P (aji, ?/i, z^) on the sur
face is connected with the origin, then the line OP lies entirely
on the surface. For, (Xaji, Xa:2, X%) are the coordinates of any
point on this line (see § 331), and they arc seen to satisfy the
given equation (9), for all values of X.
If a = 6 the cone is a cone of revolution.
EXERCISES
1. Construct and discuss each of the following surfaces.
(a) x2 + t/2;22 = o. (6)9x2 + 4 2/236020. (c) x2y2 4.4;s2 = o.
2. A point P moves so as to be equidistant from a plane and a line
perpendicular to the plane. Find the equation of the locus of P.
3. A point P moves so that the sum of its distances from the three
coordinate planes is equal to its distance from the origin. Find the
equation of the locus of P.
XXII, § 350] QUADRIC SURFACES 529
350. Summary. The surfaces discussed are here enumer
ated for reference.
Ellipsoid :
1 + !i + 2=l (Figs. 265, 266, § 343)
w 0^ c^
HyPERBOLOID OF ONE SHEET:
! + S  ^ = ^ (^^^« 2^^' 269, § 345)
0} ©2 Qi
Hyperboloid of two sheets :
a;2 ?/2
1. (Figs. 270, 271, § 346)
Elliptic Paraboloid :
a2 62
Hyperbolic Paraboloid :
/2
t I 1 = z. (Figs. 272, 273, § 347)
^^ = ^. (Figs. 274, 275, § 348)
a
Quadric Cone:
S + S = 0 (Kg. 276, § 349)
QuADRic Cylinders :
a;2
±2 = 1, y^ = ^px, (§340)
It is beyond the scope of this book to prove that the general
equation of the second degree in three variables x, y, z, can, in
general, be reduced to one of the above types. Those inter
ested in this problem will find it fully discussed in any stand
ard textbook on solid analytic geometry.*
*See, for example, Snyder and Sisam, Analytic Oeometry of Space,
Chapter 7.
2m
530
MATHEMATICAL ANALYSIS [XXII, § 351
y
351. Other Systems of Coordinates.
Numerous systems of coordinates for deter
*'^ mining the position of a point P in space have
been devised. The most common of these sys
• ^^ tems are the rectangular, polar, spherical, and
cylindrical. A brief account of the last three systems follows.
352. Polar Coordinates. Consider the line OP drawn from
the origin to any point P (Fig. 277). Let a, y8, y be the
direction angles of OP, called the radius vector, and let p be the
length of the radius vector. The four quantities a, p, y, p are
called the polar coordinates of P.
Conversely, any four quantities a, fi, y, p, with the restric
tion that cos2 a 4 cos^ ^ ) cos^ y = 1, determine a point whose
polar coordinates are a, p, y, p.
Prove that tlie equations of transformation from rectangular to polar
coordinates are.
(10)
x = p cos a, y = p cos p, z = p cos 7, p2 = x^^ +y^ h z^
353. Spherical Coordinates. Any point P in space de
termines (Fig. 278) the radius vector OP(=p), the angle <f>
between the radius vector and the 2!axis, and
the angle 6 between the a>axis and the pro
jection of the radius vector on the a;2^plane.
The quantities p, 0, <f> are called the spherical
coordinates of the point P. The angle <^ is
known as the colatitude, and the angle 6 as
the longitude.
Conversely, any three quantities p, 6, <f> determine in space a
point P whose spherical coordinates are p, 6, <i>.
Prove that the equations of transformation from rectangular to spheri
cal coordinates are,
(11) X = pBind coB(p, y = pain6sm(f>, z — p cos 6.
Fia. 278
XXII, § 354] QUADRIC SURFACES 531
354. Cylindrical Coordinates. Any point P in space de
termines (Fig. 279) its distance z from the ^p
ajyplane and the polar coordinates r, 6 of
the point P' which is the projection of P on
the a;?/plane. These three quantities r, 0, z
are called the cylindrical coordinates of P.
Conversely, any three quantities r, B, z deter
mine a point whose cylindrical coordinates they are.
Prove that the equations of transformations from rectangular to cylin
drical coordinates are,
(12) x = rQ.os,e, ?/ = rsin^, z = z.
EXERCISES
1. Express each of the following loci in spherical coordinates.
(a) a;2 I ?/2 + z^ = 9. (6) a;2 _. ^a _ 4^2 _ 0. (c) 4x2 + Qy'2  z^  36.
2. Express each of the following loci in polar coordinates.
(a) ic2 + 2/2 4. 2;2 = 16. (6) X + 1/ = 0. (c) 2 x^  y'^  ^^ = 0.
3. Express each of the following loci in cylindrical coordinates,
(a) x^ + y'^ = 9. (6) x^ + y^ + z^ = 9. (c) z'^  x'^ \ y^ = 6.
4. Express the distance between two points in polar coordinates.
5. Find the polar, spherical, and cylindrical coordinates of the points
whose rectangular coordinates are (2, 1, 4), (3, 3, 3).
6. What is the locus of points for which
(a) ^ = a constant, = a constant (spherical coordinates) ?
(6) r = a constant, 6 z= sl constant (cylindrical coordinates) ?
7. Find the general equation of a plane in polar coordinates.
8. Find the general equation of a plane in spherical coordinates;
in cylindrical coordinates.
9. Show that in polar co5rdinates a point may be regarded as the
intersection of a sphere and three cones of revolution which have an
element in common.
10. Show that in spherical coordinates a point may be regarded as
the intersection of a sphere, a plane, and a cone of revolution which are
mutually perpendicular.
11. The spherical coordinates of a point are 5, 7r/4, 7r/6; find its
rectangular coordinates ; its polar coordinates; its cylindrical coordinates.
TABLES
TO
FOUE DECIMAL PLACES
534
Powers and Roots
Squares and Cubes Square Roots and Cube Roots
No.
Sqitaee
OUBK
Square
Root
Cube
Root
No.
Squabe
Cttbe
Square
Root
Cube
Root
1
1
1
1.000
1.000
61
2,601
132,651
7.141
3.708
2
4
8
1.414
1.260
62
2,704
140,608
7.211
3.733
3
9
27
1.732
1.442
63
2,809
148,877
7.280
3.756
4
16
64
2.000
1.587
64
2,916
157,464
7.348
3.780
5
25
125
2.236
1.710
55
3,025
l(i6,375
7.416
3.803
6
36
216
2.449
1.817
56
3,136
175,616
7.483
3.826
7
49
343
2.646
1.913
57
3,249
185,193
7.550
3.849
8
64
512
2.828
2.000
58
3,364
195,112
7.616
3 871
9
81
729
3.000
2.080
59
3,481
205,379
7.681
3.893
10
100
1,000
3.162
2.154
60
3,600
216,000
7.746
3.915
11
121
1,331
3.317
2.224
61
3,721
226,981
7.810
3.936
12
144
1,728
3.464
2.289
62
3,844
238,328
7.874
3.958
13
169
2,197
3.606
2.351
63
3,969
250,047
7.937
3.979
14
196
2,744
3.742
2.410
64
4,096
262,144
8.000
4.000
15
225
3,375
3.873
2.466
65
4,225
274,625
8.062
4.021
16
256
4,096
4.000
2.520
66
4,356
287,496
8.124
4.041
17
289
4,913
4.123
2.571
67
4,489
300,763
8.185
4.062
18
324
5,832
4.243
2.621
68
4,624
314,432
8.246
4.082
19
361
6,859
4.359
2.668
69
4,761
328,509
8.307
4.102
20
400
8,000
4.472
2.714
70
4,900
343,000
8.367
4.121
21
441
9,261
4.583
2.759
71
5,041
357,911
8.426
4.141
22
484
10,648
4.690
2.802
72
5,184
373,248
8.485
4.160
23
529
12,167
4.796
2.844
73
5,329
389,017
8.544
4.179
24
576
13,824
4.899
2.884
74
5,476
405,224
8.602
4.198
25
625
15,625
5.000
2.924
75
5,625
421,875
8.660
4.217
26
676
17,576
5.099
2.962
76
5,776
438,976
8.718
4.236
27
729
19,683
5.196
3.000
77
5,929
456,533
8.775
4.254
28
784
21,952
5.292
3.037
78
6,084
474,552
8.832
4.273
29
841
24,389
5.385
3.072
79
6,241
493,039
8.888
4.291
30
900
27,000
6.477
3.107
80
6,400
512,000
8.944
4.309
31
961
29,791
5.568
3.141
81
6,561
531,441
9.000
4.327
32
1,024
32,768
5.657
3.175
82
6,724
551,368
9.055
4.344
33
1,089
35,937
5.745
3.208
83
6,889
571,787
9.110
4.362
34
1,156
39,304
5.831
3.240
84
7,056
592,704
9.165
4.380
35
1,225
42,875
5.916
3.271
85
7,225
614,125
9.220
4.397
36
1,296
46,656
6.000
3.302
86
7,396
636,056
9.274
4.414
37
1,369
50,653
6.083
3.332
87
7,569
658,503
9.327
4.431
38
1,444
54,872
6.164
3.362
88
7,744
681,472
9.381
4.448
39
1,521
59,319
6.245
3.391
89
7,921
704,969
9.434
4.465
40
1,600
64,000
6.325
3.420
90
8,100
729,000
9.487
4.481
41
1,681
68,921
6.403
3.448
91
8,281
753,571
9.539
4.498
42
1,764
74,088
6.481
3.476
92
8,464
778,688
9.592
4.514
43
1,849
79,507
6.557
3.503
93
8,649
804,357
9.644
4.531
44
1,936
85,184
6.633
3.530
94
8,836
830,584
9.695
4.547
45
2,025
91,125
6.708
3.557
95
9,025
857,375
9.747
4.563
46
2,116
97,336
6.782
3.583
96
9,216
884,736
9.798
4.579
47
2,209
103,823
6.856 3.609
97
9,409
912,673
9.849
4.595
48
2,304
110,592
6.928 3.634
98
9,604
941,192
9.899
4.610
49
2,401
117,649
7.000 3.659
99
9,801
970,299
9.950
4.626
60
2,500
125,000
7.071 3.684
100
10,000
1,000,000
10.000
4.642
For a more complete table, see The Macmillan Tables, pp. 94111.
Important Constants
Certain Convenient Values for n = 1 to n
535
10
71
1/n
Vn
V^
n !
l/n\
LOGlO V
1
1.000000
1.00000
1.00000
1
1.0000000
0.000000000
2
0.600000
1.41421
1.25992
2
0.5000000
0.301029996
3
0.333333
1.73205
1.44225
0.1666667
0.477121255
4
0.250000
2.00000
1.58740
24
0.0416667
0.602059991
5
0.200000
2.23607
1.70998
120
0.0083333
0.698970004
6
0.166667
2.44949
1.81712
720
0.0013889
0.778151250
7
0.142857
2.64575
1.91293
5040
0.0001984
0.845098040
8
0.125000
2.82843
2.00000
40320
0.0000248
0.903089987
9
0.111111
3.00000
2.08008
362880
0.0000028
0.954242509
10
0.100000
3.16228
2.15443
3628800
0.0000003
1.000000000
Logarithms of Important Constants
n =>= NUiMBBH
A^ALUE OF n
LoGio n
TT
3.14159265
0.49714987
liTT
0.31830989
9.50285013
7r2
9.86960440
0.99429975
V^
1.77245385
0.24857494
e = Napierian Base
2.71828183
0.43429448
M= logio e
0.43429448
9.63778431
l4i»/=logelO
2.30258509
0.36221569
180 J TT = degrees in 1 radian
57.2957795
1.75812262
TT M80 = radians in 1°
0.01745329
8.24187738
TT 4 10800 = radians in 1'
0.0002908882
6.4(i372613
TT T 648000 = radians in 1"
0.000004848136811095
4.68557487
sin 1"
0.000004848136811076
4.68557487
tan 1"
0.000004848136811152
4.68557487
centimeters in 1 ft.
30.480
1.4840158
feet in 1 cm.
0.032808
8.5159842
inches in 1 m.
39.37 (exact legal value)
1.5951654
pounds in 1 kg.
2.20462
0.3433340
kilograms in 1 lb.
0.453593
9.65666(50
g (average value)
32.16 ft./sec./sec.
1.5073
= 981 cm./sec./sec.
2.9916690
weight of 1 cu. ft. of water
62.425 lb. (max. density)
1.7953586
weight of 1 cu. ft. of air
0.0807 lb. (at 32° F.)
8.907
cu. in. in 1 (U. S.) gallon
231 (exact legal value)
2.3636120
ft. lb. per sec. in 1 H. P.
550. (exact legal value)
2.7403627
kg. m. per sec. in 1 H. P.
76.0404
1.8810445
watts in 1 H. P.
745.957
2.8727135
536
)
Four Place Logarithms
N
I
2
3
4
5
6
7
8
9
12 3
4 5 6
7 8 9
10
0000
0043
0080
0128
0170
0212
0253
02f)4
0334
0374
4 812
17 21 25
29 33 37
11
12
13
14
15
IG
17
18
19
0414
0702
1139
1461
1701
2041
2304
2553
2788
0453
0328
1173
1492
1790
2068
2330
2577
2810
0492
0864
1206
1523
1818
2095
2355
2.'>01
2833
0531
0899
1239
1553
1847
2122
2380
2625
2856
0569
0934
1271
1584
1875
2148
2405
2648
2878
0607
0969
1303
1614
1903
2175
2430
2672
2900
0645
1004
1335
1644
1931
2201
245".
2695
2923
0682
1038
1367
1673
1959
2227
2480
2718
2015
0719
1072
1399
1703
1987
2253
2504
2742
2967
0755
1106
1430
1732
2014
2279
2529
2765
2989
4 8 11
3 7 10
3 6 10
3 6 9
3 6 8
3 6 8
2 6 7
2 5 7
2 4 7
15 If, 23
14 17 21
13 16 19
12 16 18
11 14 17
11 13 16
10 12 16
9 1214
9 11 13
26 30 34
24 28 31
23 26 29
21 24 27
20 22 25
18 21 24
17 20 22
16 19 21
16 18 20
20
21
22
23
21
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
3010
3032
3054
3075
3096
3118
3139
3160
3181
3201
2 4 6
8 1113
16 17 19
3222
3424
3617
3802
3979
4150
4314
4472
4624
4771
4914
5051
5185
5315
5441
5563
5682
5798
5911
3213
3444
3636
3820
3997
4166
4330
4487
4639
4786
4928
5065
5198
5328
5453
5575
5694
5809
5922
3263
34G4
3655
3838
4014
4183
4346
4502
4651
4800
4942
5079
5211
5340
5465
5587
5705
5821
5933
3284
3483
3674
3856
4031
4200
4362
4518
4(569
3304
3502
3692
3874
4048
4216
4378
4533
4683
3324
3522
3711
3892
4065
4232
4393
4548
4698
3345
3541
3729
3909
4082
4249
4409
4564
4713
3365
3560
3747
3927
4099
4265
4425
4579
4728
3385
3579
3766
3946
4116
4281
4440
4594
4742
3404
3598
3784
3962
4133
4298
4456
4609
4757
2 4 6
2 4 6
2 4 6
2 4 5
2 4 6
2 3 6
2 3 5
2 3 6
13 4
8 10 12
8 10 12
7 9 11
7 9 11
7 9 10
7 8 10
6 8 9
6 8 9
6 7 9
14 16 18
14 16 17
13 16 17
12 14 16
12 14 16
11 13 16
11 12 14
11 12 14
10 12 13
4814
4955
5092
5224
5353
5478
5599
5717
6832
5944
4829
4843
4857
4871
4886
4900
1 3 4
6 7 9
10 11 13
4969
5105
5237
5366
5490
5611
5729
5843
6955
4983
5119
5250
5378
5502
6623
5740
5855
5966
4997
5132
6263
5391
5514
6635
6752
C866
6977
5011
6145
5276
6403
5527
6647
6763
5877
5988
6024
6159
6289
5416
5539
6658
5775
6888
5999
5038
5172
6302
6428
6551
6670
6786
5899
6010
1 3 4
1 3 4
13 4
1 2 4
12 4
12 4
12 4
1 2 3
1 2 3
6 7 8
5 7 8
5 7 8
6 6 8
6 6 7
5 6 7
5 6 7
6 6 7
4 6 7
10 11 12
911 12
9 1112
910 11
910 11
8 1011
8 9 11
8 9 10
8 910
40
6021
6031
6042
6053
6064
6075
6085
609(5
6107
6117
12 3
4 6 6
8 9 10
41
42
43
44
45
46
47
48
49
6128
6232
6335
6435
05 32
G628
(5721
6812
6902
6138
6213
6345
6444
6542
6637
6730
6821
6911
6149
6253
6355
6454
6551
664(5
6739
6830
6020
6160
6263
6365
6464
65(51
6656
6749
6839
6928
6170
6274
6375
6474
6571
6665
6758
6848
6937
6180
6284
6385
6484
6580
6675
6767
6857
6946
6191
6294
6395
6493
6590
(5684
6776
6866
6955
6201
6304
6405
6503
6599
6693
6785
6875
69(54
6212
6314
6416
6513
6609
6702
6794
6884
6972
6222
6325
6425
6622
6618
6712
6803
6893
6981
1 2 3
12 3
12 3
1 2 3
1 2 3
1 2 3
12 3
1 2 3
1 2 3
4 5 6
4 5 6
4 5 6
4 5 6
4 6 6
4 5 6
4 5 6
4 6 6
4 4 6
7 8 9
7 8 9
7 8 9
7 8 9
7 8 9
7 7 8
7 7 8
7 7 8
6 7 8
50
(5990
6998
7007
7016
7024
7033
7042
7050
7059
7067
12 3
3 4 6
6 7 8
51
52
53
54
7076
71G0
7243
7324
7084
7168
7251
7332
7093
7177
7259
7340
7101
7185
7267
7348
7110
7193
7275
7356
7118
7202
7284
7364
7126
7210
7292
7372
7135
7218
7300
7380
7143
7226
7308
7388
7152
7235
7316
739(5
1 2 3
1 2 3
12 2
1 2 2
3 4 5
3 4 5
3 4 5
3 4 6
6 7 8
6 7 7
6 6 7
6 6 7
N
1
2
3
4
5
6
7
8
9
12 2
4 5 6
7 8 9
The proportional parts are stated in full for every tenth at the righthand side.
The logarithm of any number of four significant figures can be read directly by add
ing the proportional part correspoTiding to the fourth figure to the tabular number
correspondmg to the first three figures. There may be an error of 1 in the last place.
538
Four Place Trigonometric Functions
[Characteristics of Logarith
ms orni
tted —
determine by the usual rule from the value]
Radiaks
DT^aPTTfl
Sine
Tangent
Cotangent
Cosine
i.'£*ijri&x>i:<o
Value
Logio
Value
Logio
Value
Logio
Value
Logio
.0000
.0029
0°00'
10
.0000
.0029
.0000
.0029
1.0000
.0000
90° 00'
1.5708
1.5679
.4637
.4637
343.77
.6363
1.0000
.0000
50
.0058
20
.0058
.7()48
.0058
.7648
171.89
.2352
1.0000
.0000
40
1.5650
.0087
30
.0087
.9408
.0087
.9409
114.59
.0591
1.0000
.0000
30
1.5621
.0116
40
.0116
.0658
.0116
.0658
85.940
.9342
.9999
.0000
20
1.5592
.0145
50
.0145
.1627
.0145
.1627
68.750
.8373
.9999
.0000
10
1.5563
.0175
1°00'
.0175
.2419
.0175
.2419
57.290
.7581
.9998
.9999
89° 00'
1.5533
.0204
10
.0204
.3088
.0204
.3089
49.104
.6911
.9998
.9999
50
1.5504
.0233
20
.0233
.3668
.0233
.3669
42.964
.6331
.9997
.9999
40
1.547..
.0202
30
.0262
.4179
.02()2
.4181
38.188
.5819
.99f)7
.9999
30
1.5440
.0291
40
.0291
.4637
.0291
.4638
34.368
.5362
.9996
.9998
20
1.5417
.0320
50
.0320
.5050
.0320
.5053
31.242
.4947
.9995
.9998
10
1.5388
.0349
2° 00'
.0349
.5428
.0349
.5431
28.636
.4569
.9994
.99f)7
88° 00'
1.5359
.0.378
10
.0378
.5776
.0378
.5779
26.432
.4221
.9993
.9997
50
1.5330
.0407
20
.0407
.6097
.0407
.6101
24.542
.3899
.9992
.9996
40
1.5301
.0436
30
.0436
.6397
.0437
.6401
22.904
.3599
.99{)0
.9996
30
1.5272
.0465
40
.0465
.6677
.0466
.6682
21.470
.3318
.9989
.9995
20
1.5243
.0495
50
.0494
.6940
.0495
.6945
20.20(5
.3055
.t)988
.9995
10
1.5213
.0524
3° 00'
.0523
.7188
.0524
.7194
19.081
.2806
.9986
.9994
87° 00'
1.5184
.0553
10
.0552
.7423
.0553
.7429
18.075
.2571
.9985
.9993
50
1.5155
.0582
20
.0581
.7645
.0582
.7652
17.169
.2348
.9983
.9993
40
1.5126
.0611
30
.0610
.7857
.0612
.7865
16..350
.2135
.9981
.9992
30
1.5097
.0640
40
.0640
.8059
.0641
.8067
15.605
.1933
.9980
.9991
20
1.5068
.0669
50
.0669
.8251
.0070
.8261
14.924
.1739
.9978
.9990
10
1.5039
.0698
4° 00'
.0698
.8436
.0699
.8446
14.301
.1554
.9976
.9989
86° 00'
1.5010
.0727
10
.0727
.8613
.0729
.8624
13.727
.1376
.9974
.9989
50
1.4981
.0756
20
.0756
.8783
.0758
.8795
13.197
.1205
.9971
.9988
40
1.4952
.0785
30
.0785
.8946
.0787
.8960
12.706
.1040
.9969
.9987
30
1.4923
.0814
40
.0814
.9104
.0816
.9118
12.251
.0882
.9967
.9986
20
1.4893
.0844
50
.0843
.9256
.0846
.9272
11.826
.0728
.9964
.9985
10
1.4864
.0873
6° 00'
.0872
.9403
.0875
.9420
11.430
.0580
.9962
.9983
86° 00'
1.4835
.0902
10
.0901
.9545
.0904
.9563
11.059
.0437
.9959
.9982
60
1.4806
.0931
20
.0929
.9682
.0934
.9701
10.712
.0299
.9957
.9981
40
1.4777
.0960
30
.0958
.9816
.0963
.9836
10.385
.0164
.9954
.9980
30
1.4748
.0985)
40
.0987
.9945
.0992
.9966
10.078
.0034
.9951
.9979
20
1.4719
.1018
50
.1016
.0070
.1022
.0093
9.7882
.9907
.9948
.9977
10
1.4690
.1047
6° 00'
.1045
.0192
.1051
.0216
9.5144
.9784
.9945
.9976
84° 00'
1.4661
.1076
10
.1074
.0311
.1080
.0336
9.2553
.9664
.9942
.9975
50
1.4632
.1105
20
.1103
.0426
.1110
.0453
9.0098
.9547
.9939
.9973
40
1.4603
.1134
30
.1132
.0539
.1139
.0567
8.7769
.9433
.993(5
.9972
30
1.4573
.1164
40
.1161
.0648
.1169
.0678
8.5555
.9322
.<)i)32
.9971
20
1.4544
.1193
50
.1190
.0755
.1198
.0786
8.3450
.9214
.9929
.9909
10
1.4615
.1222
7° 00'
.1219
.0859
.1228
.0891
8.1443
.9109
.9925
.9968
83° 00'
1.4486
.1251
10
.1248
.0961
.1257
.0995
7.9530
.JK)05
M^2
.9966
60
1.4457
.1280
20
.1276
.1060
.1287
.109(i
7.7704
.8904
.9918
.9964
40
1.4428
.1309
30
.1305
.1157
.1317
.1194
7.5958
.880(i
.9914
.9^)63
30
1.4399
.1338
40
.1334
.1252
.1346
.1291
7.4287
.8709
.9911
.9961
20
1.4370
.1367
50
.1363
.1345
.1376
.1385
7.2(i87
.8615
.9907
.9959
10
1.4341
.1396
8° 00'
.1392
.14.36
.1405
.1478
7.1154
.8522
.9903
.9958
82° 00'
1.4312
.1425
10
.1421
.1525
.1435
.1.569
6.9682
.8431
.9899
.9<)56
50
1.4283
.1454
20
.1449
.1612
.1465
.1658
6.8269
.8342
.9894
.9954
40
1.4254
.148'1
30
.1478
.1697
1495
.1745
6.6912
.8255
.9890
.9952
30
1.4224
.1513
40
.1507
.1781 .1524
.18.31
().5606
.8169
.9886
.9950
20
1.41^)5
.1542
50
.1536
.1863 .1.5.54
.1915
6.4348
.8085
.9881
.9948
10
1.4166
.1571
9° 00'
.1564
.1943
.1584
.1997
6.3138
.8003
.9877
.9946
81° 00'
1.4137
V^alne
Logjo
Value
Lopin
Value
Lopio
Value
Logio
Degrees
Rapians
Cosine
Cotangent
Tangent
Sink
Four Place Trigonometric Functions
539
[Charactoristics of Logarithms omitted —
letermine by the usual rule from the value]
Radians
Degeees
Sine
Tangent
Cotangent
Cosine
7alue
Logio
Value
Logio
Value
Logio
V^alue
Logio
.1571
9° 00'
.1564
.1943
.1584
.1997
6.3138
.8003
.9877
.9946
81° 00'
1.4137
.1600
10
.1593
.2022
.1614
.2078
6.1970
.7922
.9872
.9944
50
1.4108
.1029
20
.1622
.2100
.1044
.2158
6.0844
.7842
.98(38
.9942
40
1.4079
.1058
30
.1650
.2176
.1073
.2236
5.9758
.7764
.9863
.9940
30
1.4050
.1087
40
.1079
.2251
.1703
.2313
5.8708
.7687
.9858
.9938
20
1.4021
.1716
50
.1708
.2324
.1733
.2389
5.7694
.7611
.9853
.9936
10
1.3992
.1745
10° 00'
.1736
.2397
.1763
.2463
5.6713
.7537
.9848
.9934
80° 00'
1.3963
.1774
10
.1765
.2468
.1793
.2536
5.5764
.7464
.9843
.9931
50
1.3934
.1804
20
.1794
.2538
.1823
.2609
5.4845
.7391
.9838
.9929
40
1.3904
.1833
30
.1822
.2600
.1853
.2680
5.3955
.7320
.9833
.992;
30
1.3875
.1802
40
.1851
.2674
.1883
.2750
5.3093
.7250
.9827
.9924
20
1.3840
.1891
50
.1880
.2740
.1914
.2819
5.2257
.7181
.9822
.9922
10
1.3817
.1920
11°00'
.1908
.2806
.1944
.2887
5.1446
.7113
.9816
.9919
79° 00'
1.3788
.1949
10
.1937
.2870
.1974
.2953
5.0058
.7047
.9811
.9917
50
1.3759
.1978
20
.1965
.2934
.2004
.3020
4.9894
.6980
.9805
.9914
40
1.3730
.2007
30
.1994
.25)97
.2035
.3085
4.9152
.6915
.9799
.9912
30
1.3701
.2036
40
.2022
.3058
.2065
.3149
4.8430
.6851
.9793
.9909
20
1.3672
.2065
50
.2051
.3119
.2095
.3212
4.7729
.6788
.9787
.9907
10
1.3&43
.2094
12° 00'
.2079
.3179
.2126
.3275
4.7046
.6725
.9781
.9904
78° 00'
1.3014
.2123
10
.2108
.3238
.2156
.3330
4.6382
.6664
.9775
.9901
■50
1.3584
.2153
20
.2136
.3296
.2186
.3397
4.5736
.6603
.9769
.9899
40
1.3555
.2182
30
.2164
.3353
.2217
.3458
4.5107
.6542
.9763
.9896
30
1.3526
.2211
40
.2193
.3410
.2247
.3517
4.4494
.6483
.9757
.9893
20
1.3497
.2240
60
.2221
.3466
.2278
.3576
4.3897
.6424
.9750
.9890
10
1.3468
.2269
13° 00'
.2250
.3521
.2309
.3634
4.3315
.636()
.9744
.9887
77° 00'
1.3439
.2298
10
.2278
.3575
.2339
.3691
4.2747
.6309
.9737
.9884
50
1.3410
.2327
20
.2306
.3629
.2370
.3748
4.2193
.0252
.9730
.9881
40
1.3381
.2356
30
.2334
.3682
.2401
.3804
4.1653
.6196
.9724
.9878
30
1.3352
.2385
40
.2363
.3734
.2432
.3859
4.1126
.6141
.9717
.9875
20
1.3323
.2414
60
.2391
.3786
.2462
.3914
4.0611
.6080
.9710
.9872
10
1.3294
.2443
14° 00'
.2419
.3837
.2493
.3968
4.0108
.6032
.9703
.9869
76° 00'
1.3265
.2473
10
.2447
.3887
.2524
.4021
3.9017
.5979
.96%
.9866
50
1.3235
.2502
20
.2476
.3937
.2555
.4074
3.9136
.5926
.9689
.9863
40
1.3206
.2531
30
.2504
.3986
.2586
.4127
3.8667
.5873
.9681
.9859
30
1.3177
.2560
40
.2532
.4035
.2617
.4178
3.8208
.5822
.9674
.9856
20
1.3148
.2589
50
.2560
.4083
.2648
.4230
3.7760
.5770
.9(367
.9853
10
1.3119
.2618
15°00'
.2588
.4130
.2679
.4281
3.7321
.5719
.9659
.9849
75° 00'
1.3090
.2647
10
.2616
.4177
.2711
.4331
3.0891
.5669
.9652
.9846
50
1.3061
.2676
20
.2644
.4223
.2742
.4381
3.6470
.5619
.9644
.9843
40
1.3032
.2705
30
.2672
.4269
.2773
.4430
3.6059
.5570
.9(>36
.9839
30
1.3003
.2734
40
.2700
.4314
.2805
.4479
3.5656
.5521
.9(528
.9836
20
1.2974
.2763
60
.2728
.4359
.2836
.4527
3.5261
.5473
.9621
.9832
10
1.2945
.2793
16° 00'
.2756
.4403
.2867
.4575
3.4874
.5425
.9613
.9828
74° 00'
1.2915
.2822
10
.2784
.4447
.2899
.4622
3.4495
.5378
.9605
.9825
50
1.2886
.2851
20
.2812
.4491
.2931
.4669
3.4124
.5331
.9596
.9821
40
1.2857
.2880
30
.2840
.4533
.2962
.4716
3.3759
.5284
.9588
.9817
30
1.2828
.2909
40
.2868
.4576
.2994
.4762
3.3402
.5238
.9580
.9814
20
1.2799
.2938
50
.2896
.4618
.3026
.4808
3.3052
.5192
.9572
.9810
10
1.2770
.2967
17° 00'
.2924
.4659
.3057
.4853
3.2709
.5147
.9563
.9806
73° 00'
1.2741
.2996
10
.2952
.4700
.3089
.4898
3.2371
.5102
.9555
.9802
50
1.2712
.3025
20
.2979
.4741
.3121
.4943
3.2041
.5057
.9546
.9798
40
1.2683
.3054
30
.3007
.4781
.3153
.4987
3.1716
.5013
.9537
.9794
30
1.2654
.3083
40
.3035
.4821
.3185
.5031
3.1397
.4969
.9528
.9790
20
1.2625
.3113
50
.3062
.4861
.3217
.5075
3.1084
.4925
.9520
.9786
10
1.2696
.3142
18^00'
.3090
.4900
.3249
.5118
3.0777
.4882
.9511
.9782
72° 00'
1.2566
Value
LoSio
Value
Loffio
Value
T^'^ffio
Value
Logjo
Degeees
Radians
Co'JINE
Cotangent
Tangent
Sine
540
Four Place Trigonometric Functions
[Characteristics of Logarithms omitted —
ietermine by the usual rule from the value]
Kadians
Deobees
Sine
Tangent
Cotangent
Cosine
Value
Logio
Value
Logio
Value
Logio
Value
Lopio
.3142
18° 00'
.3090
.4900
.3249
.5118
3.0777
.4882
.9511
.9782
72° 00'
1.2566
.3171
10
.3118
.4939
.3281
.6161
3.0475
.4839
.9502
.9778
60
1.2537
.3200
20
.3145
.4977
.3314
.6203
3.0178
.4797
.9492
.9774
40
1.2508
.3229
30
.3173
.5015
.3346
.6245
2.9887
.4756
.9483
.9770
30
1.2479
.3258
40
.3201
.5052
.3378
.6287
2.9600
.4713
.9474
.9765
20
1.2450
.3287
50
.3228
.5090
.3411
.5329
2.9319
.4671
.9465
.9761
10
1.2421
.3316
19° 00'
.3256
.5126
.3443
.5370
2.9042
.4630
.9455
.9767
71° 00'
1.2392
.3346
10
.3283
.5163
.3476
.5411
2.8770
.4689
.9446
.9752
60
1.2363
.3374
20
.3311
.5199
.3508
.5451
2.8502
.4649
.9436
.9748
40
1.2334
.3403
30
.3338
.5235
.3541
.5491
2.8239
.4509
.9426
.9743
30
1.2305
.3432
40
.3365
.5270
.3574
.6531
2.7980
.4469
.9417
.9739
20
1.2275
.3462
50
.3393
.5306
.3607
.5571
2.7725
.4429
.9407
.9734
10
1.2246
.3491
20° 00'
.3420
.5341
.3640
.6611
2.7476
.4389
.9397
.9730
70° 00'
1.2217
.3520
10
.3448
.5375
.3673
.5650
2.7228
.4350
.9387
.9725
50
1.2188
.3549
20
.3475
.5409
.3706
.5689
2.6986
.4311
.9377
.9721
40
1.2159
.3578
30
.3502
.5443
.3739
.6727
2.6746
.4273
.9367
.9716
30
1.2130
.3607
40
.3529
.5477
.3772
.6766
2.6511
.4234
.9356
.9711
20
1.2101
.3636
50
.3557
.5510
.3805
.5804
2.6279
.4196
.9346
.9706
10
1.2072
.3665
21° 00'
.3584
.5643
.3839
.5842
2.6051
.4158
.9336
,9702
69° 00'
1.2043
.3694
• 10
.3611
.5576
.3872
.6879
2.5826
.4121
.9325
.9697
60
1.2014
.3723
20
.3638
.5609
.3906
.5917
2.5605
.4083
.9315
.9692
40
1.1985
.3752
30
.3m5
.5641
.3939
.5954
2.5386
.4046
.9304
.9687
30
1.1956
.3782
40
.3692
.5673
.3973
.5991
2.5172
.4009
.9293
.9682
20
1.1926
.3811
50
.3719
.5704
.4006
.6028
2.4960
.3972
.9283
.9677
10
1.1897
.3840
22° 00'
.3746
.6736
.4040
.6064
2.4751
.3936
.9272
.9672
68° 00'
1.1868
.3869
10
.3773
.5767
.4074
.6100
2.4545
.3900
.9261
.9667
50
1.1839
.3898
20
.3800
.5798
.4108
.6136
2.4342
.3864
.9250
.9661
40
1.1810
.3927
30
.3827
.6828
.4142
.6172
2.4142
.3828
.9239
.9(556
30
1.1781
.3956
40
.3854
.5859
.4176
.6208
2.3945
.3792
.9228
.9651
20
1.1752
.3985
50
.3881
.5889
.4210
.6243
2.3750
.3767
.9216
.9646
10
1.1723
.4014
23° 00'
.3907
.6919
.4245
.6279
2.3559
.3721
.9205
.9640
67° 00'
1.1694
.4043
10
.3934
.5948
.4279
.6314
2.3309
.3686
.9194
.9635
50
1.1(>65
.4072
20
.3961
.5978
.4314
.6348
2,3183
.3652
.9182
.9629
40
1.1636
.4102
30
.3987
.6007
.4348
.6383
2.2998
.3617
.9171
.9624
30
1.160(5
.4131
40
.4014
.6036
.4383
.6417
2.2817
.3583
.9159
.9618
20
1.1577
.4160
60
.4041
.6065
.4417
.6462
2.2637
.3548
.9147
.9613
10
1.1548
.4189
24° 00'
.4067
.6093
.4452
.6486
2.24(50
.3514
.9135
.9607
66° 00'
1.1519
.4218
10
.4094
.6121
.4487
.6520
2.2286
.3480
.9124
.9602
50
1.1490
.4247
20
.4120
.6149
.4522
.(5553
2.2113
.3447
.9112
.9596
40
1.1461
.4276
30
.4147
.6177
.4557
.6587
2.1943
.3413
.9100
.9590
30
1.1432
.4305
40
.4173
.6206
.4592
.6620
2.1775
.3380
.9088
.9584
20
1.1403
.4334
50
.4200
.6232
.4628
.6654
2.1609
.3346
.9075
.9579
10
1.1374
.4363
25° 00'
.4226
.6259
.4663
.6687
2.1445
.3313
.9063
.9573
65° 00'
1.1345
.4392
10
.4253
.6286
.4699
.6720
2.1283
.3280
.<X)51
.95(57
50
1.1316
.4422
20
.4279
.6313
.4734
.6752
2.1123
.3248
.9038
.9561
40
1.128(5
.4451
30
.4305
.6340
.4770
.6785
2.0965
.3215
.9026
.95,^5
30
1.1257
.4480
40
.4331
.6.366
.4806
.6817
2.0809
.3183
.9013
.9549
20
1.1228
.4509
50
.4358
.6392
.4841
.6850
2.0656
.3150
.9001
.9543
10
1.1199
.4538
26° 00'
.4384
.6418
.4877
.6882
2.0503
.3118
.8988
.9537
64° 00'
1.1170
.4567
10
.4410
.6444
.4913
.6914
2.0353
.3086
.8975
.9.530
50
1.1141
.4596
20
.4436
.6470
.4950
.6946
2.0204
.3054
.89(52
.9624
40
1.1112
.4625
30
.4462
.0495
.4986
.(5977
2.0057
.3023
.8949
.9518
30
1.1083
.4654
40
.4488
.6621
.5022
.7009
1.9^)12
.2991
.89:^5
.9512
20
1.1054
.4083
60
.4514
.6646
.5059
.7040
1.9768
.2960
.8923
.9505
10
1.1025
.4712
27° 00'
.4540
.6670
.6096
.7072
1.9626
.2928
.8910
.9499
63° 00'
1.0996
Value
Logio
Value
Lopio
Value
Logio
Value
Log,o
Degrees
Radians
Cosine
Cotangent
Tangent
Sine
Four Place Trigonometric Functions 541
[Oharaoteristics of Logarithms omitted — determine by the usual rule from the value]
Radians
Dbgebes
SI.B
Tangent
Cotangent
Cosine
Value
Loffio
Value
Logio
Value Logio
Value
Logio
.4712
27° 00'
.4540
.6570
.5095
.7072
1.9626 .2923
.8910
.9499
63° 00'
1.0996
.4741
10
.4566
.6595
.5132
.7103
1.9486 .2897
.8897
.9492
50
1.0966
.4771
20
.4592
.6620
.5169
.7134
1.9347 .2866
.8884
.948(5
40
1.0937
.4800
30
.4617
.6644
.5206
.7165
1.9210 .2835
.8870
.9479
30
1.0908
.4829
40
.4643
.6663
.5243
.7196
1.9074 .2804
.8857
.9473
20
1.0879
.4858
60
.4669
.6692
.5280
.7226
1.8940 .2774
.8843
.9466
10
1.0850
.4887
28° 00'
.4695
.6716
.5317
.7257
1.8807 .2743
.8829
.9459
62° 00'
1.0821
.4Q16
10
.4720
.6740
.5354
.7287
1.8676 .2713
.8816
.9453
50
1.0792
.4945
20
.4746
.6763
.5392
.7317
1.8546 .2683
.8802
.9446
40
1.0703
.4974
30
.4772
.6787
.5430
.7348
1.8418 .2652
.8788
.9439
30
1.0734
.5003
40
.4797
.6810
.5167
.7378
1.8291 .2(522
.8774
.9432
20
1.0705
.5032
50
.4823
.6833
.5505
.7408
1.8165 .2592
.8760
.9425
10
1.0676
.5061
29° 00'
.4848
.6856
.5543
.7438
1.8040 .2562
.8746
.9418
61° 00'
1.0647
.5091
10
.4874
.6878
.5581
.7467
1.7917 .2533
.8732
.9411
50
1.0617
.5120
20
.4899
.6901
.5619
.7497
1.7796 .2503
.8718
.9404
40
1.0588
.5149
30
.4924
.6923
.5658
.7526
1.7675 .2474
.8704
.9397
30
1.0559
.5178
40
.4950
.6946
.5696
.7556
1.7556 .2444
.8689
.9390
20
1.0530
.5207
50
.4975
.6968
.5735
.7585
1.7437 .2415
.8675
.9383
10
1.0501
.5236
30° 00'
.5000
.6990
.5774
.761i
1.7321 .2386
.8660
.9375
60° 00'
1.0472
.5265
10
.5025
.7012
.7(544
1.7205 .2356
.8646
.9368
50
1.0443
.5294
20
.50jO
.7033
.5851
.7673
1.7090 .2327
.8631
.9361
40
1.0414
.5323
30
.5075
.7055
.5890
.7701
1.6977 .2299
.8616
.9353
30
1.0385
.5352
40
.5100
.7076
.5930
.7730
1.6864 .2270
.8601
.9346
20
1.0356
.5381
50
.5125
.7097
.5969
.7759
1.6753 .2241
.8587
.9338
10
1.0327
.5411
31°00'
.5150
.7118
.6009
.7788
1.6643 .2212
.8572
.9331
59° 00'
1.0297
.5440
10
.5175
.7139
.6048
.7816
1.6534 .2184
.8557
.9323
.50
1.0268
.5469
20
.5200
.7160
.6088
.7845
1.6426 .2155
.8542
.9315
40
1.0239
.5498
30
.5225
.7181
.6128
.7873
1.6319 .2127
.8526
.9308
30
1.0210
.5527
40
.5250
.7201
.6168
.7902
1.6212 .2098
.8511
.9300
20
1.0181
.5556
50
.5275
.7222
.6208
.7930
1.6107 .2070
.8496
.9292
10
1.0152
.5585
32° 00'
.5299
.7242
.6249
.7958
1.6003 .2042
.8480
.9284
58° 00'
1.0123
.5614
10
.5324
.7262
.6289
.7986
1.5900 .2014
.8465
.9276
50
1.0094
.5643
20
.5348
.7282
.6330
.8014
1.5798 .1986
.8450
.9268
40
1.0065
.5672
30
.5373
.7302
.6371
.8042
1.5697 .1958
.8434
.9260
30
1.0036
.5701
40
.5398
.7322
.6412
.8070
1.5597 .1930
.8418
.9252
20
1.0007
.5730
50
.5422
.7342
.6453
.8097
1.5497 .1903
.8403
.9244
10
.9977
.5760
33° 00'
.5446
.7361
.6494
.8125
1.5399 .1875
.8387
.9236
57° 00'
.9948
.5789
10
.5471
.7380
.6536
.8153
1.5301 .1847
.8371
.9228
50
.9919
.5818
20
.5495
.7400
.6577
.8180
1.5204 .1820
.8.355
.9219
40
.9890
.5847
30
.5519
.7419
.6619
.8208
1.5108 .1792
.8339
.9211
30
.9861
.587(5
40
.5544
.7438
.6(561
.8235
1.5013 .1765
.8323
.9203
20
.9832
.5905
50
.5568
.7457
.6703
.8263
1.4919 .1737
.8307
.9194
10
.9803
.5934
34° 00'
.5592
.7476
.6745
.8290
1.4826 .1710
.8290
.9186
66° 00'
.9774
.5903
10
.5616
.7494
.6787
.8317
1.4733 .1683
.8274
.9177
50
.9745
.5992
20
.5640
.7513
.6830
.8344
1.4641 .1656
.8258
.9169
40
.9716
.6021
30
.5664
.7531
.6873
.8371
1.4550 .1629
.8241
.9160
30
.9687
.6050
40
.5688
.7550
.6916
.8398
1.4460 .1602
.8225
.9151
20
.9657
.6080
50
.5712
.7568
.6959
.8425
1.4370 .1575
.8208
.9142
10
.9628
.6109
35° 00'
.5736
.7586
.7002
.8452
1.4281 .1548
.8192
.9134
55° 00'
.9599
.6138
10
.5760
.7604
.7046
.8479
1.4193 .1521
.8175
.9125
50
.9570
.6167
20
.5783
.7622
.7089
.8506
1.4106 .1494
.8158
.9116
40
.9541
.6196
30
.5807
.7640
.7133
.8533
1.4019 .1467
.8141
.9107
30
.9512
.6225
40
.5831
.7657
.7177
.8559
1.3934 .1441
.8124
.9098
20
.9483
.6254
50
.5854
.7675
.7221
.8586
1.3848 .1414
.8107
.9089
10
.9454
.6283
36° 00'
.5878
.7692
.7265
.8613
1.3764 .1387
.8090
.9080
54° 00'
.9425
Value
Logio
Value
Loffio
Value Logio
Value
I^Ogio
Degeees
Radians
Cosine
Cotangent
Tangent
Sine
542 Four Place Trigonometric functions
[Characteristics of Logarithms omitted — determine by the usual rule from the value]
Radians
Deobeee
8lNB
Tangent
Cotangent
Cosine
Value Logio
Value Logio
Value Logifl
Value Logic
.6283
36° 00'
.5878 .7692
.7265 .8613
1.3764 .1387
.8090 .9080
64° 00'
.9425
.6332
10
.5901 .7710
.7310 .8639
1.3680 .1361
.8073 .9070
60
.9396
.6341
20
.5925 .7727
.7355 .86(J6
1.3597 .1334
.8056 .9061
40
.9367
.6370
30
.5948 .7744
.7400 .8692
1.3514 .1308
.8039 .9052
30
.9338
.6400
40
.5972 .7761
.7445 .8718
1.3432 .1282
.8021 .9042
20
.9308
.6429
50
.5995 .7778
.7490 .8745
1.3351 .1255
.8004 .9033
10
.9279
.6458
37° 00'
.6018 .7795
.7536 .8771
1.3270 .1229
.7986 .9023
63° 00'
.9250
.6487
10
.6041 .7811
.7581 .8797
1.3190 .1203
.79f39 .tX)14
50
.9221
.6516
20
.()065 .7828
.7627 .8824
1.3111 .1176
.7951 .9004
40
.9192
.6545
30
.6088 .7844
.7()73 .8850
1.3032 .1150
.7934 .8995
30
.9163
.6574
40
.6111 .7861
.7720 .8876
1.2954 .1124
.7916 .8985
20
.9134
.6603
50
.6134 .7877
.77(J6 .8902
1.2876 .1098
.7898 .8975
10
.9105
.6632
38° 00'
.6157 .7893
.7813 .8928
1.2799 .1072
.7880 .8965
52° 00'
.9076
.6601
10
.6180 .7910
.7860 .8954
1.2723 .1046
.7862 .8955
50
.9047
.6u90
20
.6202 .7926
.7907 .8980
1.2647 .1020
.7844 .8945
40
.9018
.6720
30
.6225 .7941
.79r>4 .9006
1.2572 .0994
.7826 .8935
30
.8988
.6749
40
.6248 .7957
.8002 .9032
1.2497 .0968
.7808 .8925
20
.8959
.6778
50
.6271 .7973
.8050 .9058
1.2423 .0942
.lim .8915
10
.8930
.6807
39° 00'
.6293 .7989
.8098 .9084
1.2349 .0916
.7771 .8905
61° 00'
.8901
.6836
10
.6316 .8004
.8146 .9110
1.2276 .0890
.7753 .8895
50
.8872
.6865
20
.6338 .8020
.8195 .9135
1.2203 .0865
.7735 .8884
40
.8843
.6894
30
.6361 .8035
.8243 .9161
1.2131 .0839
.7716 .8874
30
.8814
.6923
40
.6383 .8050
.8292 .9187
1.2059 .0813
.7698 .8864
20
.8785
.6952
60
.6406 .8066
.8342 .9212
1.1988 .0788
.7679 .8853
10
.8756
.6981
40° 00'
.6428 .8081
.8391 .9238
1.1918 .0762
.7660 .8843
60° 00'
.8727
.7010
. 10
.6450 .8096
.8441 .9264
1.1847 .0736
.7642 .8832
50
.8698
.7039
20
.6472 .8111
.8491 .9289
1.1778 .0711
.7623 .8821
40
.8668
.7069
30
.6494 .8125
.8541 .9315
1.1708 .0685
.7604 .8810
30
.8639
.7098
40
.6517 .8140
.8591 .9341
1.1640 .0659
.7585 .8800
20
.8610
.7127
50
.6539 .8155
.8642 .9366
1.1571 .0634
.7566 .8789
10
.8581
.7156
41° 00'
.6561 .8169
.8693 .9392
1.1504 .0608
.7547 .8778
49° 00'
.8552
.7185
10
.6583 .8184
.8744 .9417
1.1436 .0583
.7528 .8767
50
.8523
.7214
20
.6604 .8198
.87m .9443
1.1369 .0557
.7509 .8756
40
.8494
.7243
30
.6626 .8213
.8847 .94(38
1.1303 .0532
.7490 .8745
30
.8465
.7272
40
.6648 .8227
.8899 .9494
1.1237 .050(3
.7470 .8733
20
.8436
.7301
50
.6670 .8241
.8952 .9519
1.1171 .0481
.7451 .8722
10
.8407
.7330
42° 00'
.6691 .8255
.9004 .9544
1.1106 .0456
.7431 .8711
48° 00'
.8378
.7359
10
.6713 .8269
.9057 .9570
1.1041 .0430
.7412 .8699
50
.8348
.7389
20
.6734 .8283
.9110 .9595
1.0977 .0405
.7392 .8688
40
.8319
.7418
30
.6756 .8297
.91(33 .9()21
1.0913 .0379
.7373 .8676
30
.8290
.7447
40
.6777 .8311
.9217 .9646
1.0850 .0354
.7353 .8665
20
.8261
.7476
50
.()799 .8324
.9271 .9671
1.0786 .0329
.7333 .8(353
10
.8232
.7505
43° 00'
.6820 .8338
.9325 .9697
1.0724 .0303
.7314 .8641
47°00'
.8203
.7534
10
.6841 .8351
.9380 .9722
1.0661 .0278
.7294 .8629
60
.8174
.7563
20
.6862 .8365
.9435 .9747
1.0599 .0253
.7274 .8618
40
.8145
.7592
30
.6884 .8378
.9490 .9772
1.0538 .0228
.7254 .8606
30
.8116
.7621
40
.6905 .8391
.9545 .9798
1.0477 .0202
.7234 .8594
20
.8087
.7650
60
.6926 .8405
.9601 .9823
1.0416 .0177
.7214 .8582
10
.8058
.7679
44° 00'
.6947 .8418
.9657 .9848
1.0355 .0152
.7193 .8569
46° 00'
.8029
.7709
10
.6967 .»i31
.9713 .9874
1.0295 .012(3
.7173 .8567
60
.7999
.7738
20
.6988 .8444
.9770 .9899
1.0235 .0101
.7153 .8545
40
.7970
.7767
30
.7009 .8457
.9827 .9924
1.0176 .0076
.7133 .8532
30
.7941
.7796
40
.7030 .84(59
.9884 .9949
1.0117 .0051
.7112 .8520
20
.7912
.7825
50
.7050 .8482
.9942 .9975
1.0058 .0025
.7092 .8507
10
.7883
.7864
46° 00'
.7071 .8495
1.0000 .0000
1.0000 .0000
.7071 .849..
45° 00'
.7854
'
Value Logjo
Value Lopio
Value Logio
Value Logio
DSGSEEB
Radians
OOSINB
Cotangent
Tangent
Sine
INDEX
Abscissa, 39.
Absolute error, 236.
Absolute value, of a directed quan
tity, 5 ; of a number, 36, 438.
Addition, 41 ; graphic, 50 ; laws
of, 52 ; of angles, 145 ; formulas,
in trigonometry, 201 ; with
rounded numbers, 238 ; of vectors,
435.
Algebra, fundamental theorem of,
456.
Algebraic functions, 143.
Algebraic scale, 3, 5.
Analytic geometry, 84.
Analytic representation of a function,
21.
Angle, definitions of, 143 ; directed
— s, 143, 306 ; measurement of,
144, 188, 190; addition and
subtraction of, 145 ; functions of,
147, 168; between lines, 146
306, 500; between planes, 506
of elevation and depression, 150
use of, in artillery service, 190
vectorial, 163 ; — s of triangle,
210; trisection of, 388; of a
complex number, 438.
Approximation, Newton's method of,
471.
Arbitrary functions, 18, 19.
Arc of a circle, 189.
Archimedes, 386.
Area, of a triangle, 186, 298 fif. ; of
any polygon, 301.
Arithmetic mean, 214.
Arithmetic progression, 216.
Arithmetic scale, 3, 5.
Artillery service, use of angles in,
190.
Asymptotes, 278, 280, 350.
Axes, of reference, 38, 495 ; of ellipse,
273, 340 ; of hyperbola, 280, 349.
Axioms, 53.
Axis, of parabola, 109, 354; polar,
163 ; major and minor, 340 ; trans
verse, 280, 349 ; conjugate, 349.
Binomial theorem, 428.
Briggian logarithms, 227.
Bundle of planes, 510.
Cardioid, 383.
Center, of pencil of lines, 91 ; of
ellipse, 273, 282, 340; of circle,
283, 320 ; of hyperbola, 280, 349 ;
of projection, 370; of bundle of
planes, 510.
Change, rate of, 68.
Change ratio, 69.
Characteristic of a logarithm, 227
ff.
Circle, 320 ff . ; center and radius
of, 320 ; cartesian equation of, 270,
320 ff . ; parametric equations of,
392 ; polar equation of, 381 ; as
special case of an ellipse, 342 ;
intersection of two — s, 330 ; or
thogonal — s, 331, 334; pencil of
— s, 332 ; radical axis of, 332 ; radi
cal center of, 334 ; tangent to, point
form, 324 ; slope form, 325 ; tan
gents from an external point to,
327; polar of point with respect
to, 328 ; inversion with respect
to, 336.
Cissoid, 384.
Cof unction, 177.
Colatitude, 530.
CoUinearity, condition for, 301.
Combinations, 420 ff .
Commensurable segments, 33.
Common logarithms, 227.
Completing the square, 113, 283 ff.
543
544
INDEX
Complex numbers, 432 ff. ; defini
tion of, 432 ; geometric inter
pretation of, 433, 436 S. ; absolute
value, angle, argument of, 438 ;
polar form of, 438 ; multiplication
and division of, 440; powers and
roots of, 444.
Complex roots of an equation, 462.
Components of a vector, 436.
Composite number, 414.
Computation, numerical, 231, 236 ff.,
242 ff.
Conchoid, 385.
Cone, 528.
Conic or Conic section, 337 ff. ; as
sections of a cone, 370. (See
circle, ellipse, parabola, hyperbola.)
Conjugate axis, 349.
Conjugate complex numbers, 432.
Conjugate diameters, 376.
Conjugate hyperbolas, 353.
Consistent equations, 94, 491.
Constant function, 18.
Continuous functions, 18, 102, 449.
Coordinates, on a line, 37 ; rectan
gular, in a plane, 38 ; rectangular,
in space, 495 ; polar, in a plane,
163, 377 ff. ; polar, in space, 530 ;
spherical, 530; cylindrical, 531.
Cosecant, 168; graph of, 173.
Cosine, definition of, 147 ; variation
of, 159 ; graph of, 159 ; graph
of, in polar coordinates, 166 ; line
representation of, 169 ; law of — s,
180 ; direction — s of a line, 498.
Cotangent, definition of, 168; graph
of, 173; line representation of,
169.
Coversed sine, 168.
Cube, duplication of, 388; table of
— s, and — roots, 634.
Cubic function, 129 ff.
Cycloid, 399.
Cylinders, 515.
Cylindrical co5rdinates, 531.
Decreasing function, 24.
De Moivre's theorem, 443.
Dependent variable, 12.
Depressed equation, 469.
Derivative, of a function, 451, 468 flF. ;
successive — s, 458.
Derived function, 451.
Descartes's rule of signs, 466.
Detached coefficients, 402 ff.
Determinants, 475 ff. ; definition of,
475, 478, 483 ; evaluation of, 488 ;
properties of, 483 ff, ; minor of,
485 ; Laplace's expansion of, 486 ;
solution of equations by means of,
476, 480, 490.
Diameter, of a conic, 373 ; conjugate
— s, 376.
Diodes, 384.
Directed angles, 143, 306.
Directed lines, 30G, 500.
Directed quantities, 4.
Directed segments, 5, 48, 497.
Direction angles and cosines, 498.
Directrix, of conic, 337 ; of ellipse,
340; of hyperbola, 350; of parab
ola, 355.
Discontinuous functions, 18.
Discriminant of a quadratic, 124.
Distance, between two points in a
plane, 294 ; in space, 499 ; of a
point from a line, 311 ; of a point
from a plane, 508.
Division, by zero, 46 ; with rounded
numbers, 241 ; of complex
numbers, 440; point of, 295,
501 ; — transformation, 404.
Duplication of cube, 388.
Eccentricity, 337.
Element of a determinant, 475, 478.
Ellipse, definition of, 272, 338
axes and center of, 273, 340
equations of, 272, 282, 338 ff.
slope of, 273 ; construction of,
346 ; focal radii of, 345 ; latus
rectum of, 343 ; parametric equa
tions of, 347, 393 ; properties of,
340, 342, 365, 373; vertices,
343.
Ellipsoid, 518 ff.
Elliptic paraboloid, 526.
Empirical function, 18.
Equality, 51 ; conditional and un
conditional, 61.
INDEX
545
Equation, definition of, 61 ; linear
— s, 64, 83, 93 ff ., 509 ; quadratic,
120 £f.; trigonometric — s, 174;
exponential — s, 234 ; solution by
determinants, 476, 480, 490; in
pform, 467 ; depressed, 469 ;
parametric, 392 ff.
Error, absolute and relative, 236.
Explicit function, 81.
Exponential equations, 234.
Exponential function, 217 ff.
Exponents, 53, 218.
External secant, 168.
Factor, 51 ; of a polynomial, 404,
407; — theorem, 411.
Factoring, solution of quadratic
equation by, 121.
Fire, indirect, in artillery service,
190.
Focal radii, of ellipse, 345 ; of hyper
bola, 353.
Focus, of conic, 337 ; of ellipse, 340 ;
of hyperbola, 350; of parabola,
355.
Forces, parallelogram of, 184, 435.
Fractions, 33, 58 ; partial, 416 fif .
Function, idea of, 1 ; definition of,
28 ; arbitrary, constant, empirical,
18; continuous, 18, 102, 449;
representation of, 10, 13, 21, 22;
increasing and decreasing, 24 ;
linear, 64 ff., 494 ff. ; quadratic,
98 ff., 265 ff., 514 ff. ; cubic, 129 ff. ;
power, 140; trigonometric, 147 ff.,
168 ff.; logarithmic, 212 ff. ; ex
ponential, 217 ff. ; polynomial,
449 ff. ; explicit and implicit, 81 ;
inverse trigonometric, 192 ff. ; sum
of linear — s, 79 ; tables of — s,
534 ff. ; two valued, 20, 265 ff.
Functional notation, 409.
Fundamental theorem of algebra,
456.
Geometric mean, 214.
Geometric progression, 216.
Geometric representation, see graphic
representation.
Graphic addition, 7, 50.
2n
Graphic interpolation, 13.
Graphic representation, 3, 10, 37, 64,
433, 436 ff.
Graphic solution of problems, 78,
123, 470.
Graphs, statistical, 26 ; of linear
functions, 72 ; of quadratic func
tions, 99 ff., 265 ff. ; of cubic
functions, 129 ff. ; of polynomials,
452 ; of trigonometric functions,
158, 159, 161, 166, 173; of the
exponential function, 221 ; of the
logarithmic function, 224 ; in
polar coordinates, 164, 378 ff. ; of
parametric equations, 395. (.See
entries under various classes of
functions for further details.)
Hesse's normal form of the equation
of a straight line, 316.
Highest common factor, 407.
Hooke's law, 71.
Hyperbola, definition of, 280, 338;
axes and center, 280, 349 ; vertices
of, 349; asymptotes of, 278,
280, 355; construction of, 354;
equations of, 279, 283, 348 ff.;
parametric equations of, 393 ;
latus rectum of, 350 ; geometric
properties of, 349, 350, 368;
tangents and normals to, 359;
conjugate — s, 353.
Hyperbolic curves, 140.
Hyperbolic paraboloid, 527.
Hyperbolic spiral, 386.
Hyperboloid, of one sheet, 621 ;
of two sheets, 524.
Hypocycloid, 400.
Identities, definition of, 61 ; trigono
metric, 171.
Imaginary number, 432 ff.
Implicit functions, 81 ; quadratic
functions, 265 ff.
Incommensurable quantities, 34.
Increasing function, 24.
Independent variable, 12.
Infinity, 48. ■
Initial line, 163.
Inscribed circle, 187.
546
INDEX
Integer, 33 ; properties of, 414.
Intercept, 74, 517 ; — form of equa
tion of straight line, 315.
Interpolation, 13, 77, 230.
Intersection, of two lines, 93 ; of
two circles, 330; of a line and a
conic, 357.
Inverse trigonometric functions,
192 ff.
Inversion, with respect to a circle,
336 ; of order, 482.
Inversor of Peaucellier, 336.
Irrational numbers, 34 ; as roots of
an equation, 470.
Laplace's expansion of a determi
nant, 486.
Latus rectum, of an ellipse, 343 ; of a
hyperbola, 350 ; of a parabola, 355.
Law, of signs, 49, 466; — s of ex
ponents, 53, 220; of sines, 180/
of cosines, 180 ; of tangents, 209.
Less than, 39.
Limafon, 383.
Line representation of trigonometric
functions, 169.
Linear equations, 64, 83, 93, 476,
480, 490, 509.
Linear functions, 64 fif., 494 ff.
Linear interpolation, 77.
Locus, of equation in rectangular
coordinates, 67, 83, 509 ; in polar
coordinates, 377.
Logarithm, definition of, 223 ; in
vention of, 212; laws of, 225;
systems of (natural and common),
226, 227 ; characteristic and man
tissa of, 227 ff. ; use of tables of,
229; — s in computation, 231,
242 £F.; tables of, 536 ff.
Logarithmic paper, 260 ff.
Logarithmic scale, 252.
Logarithmic spiral, 387.
Longitude, 530.
Magnitude, 4.
Major axis of ellipse, 340.
Mathematical analysis, 30.
Maxima and Minima, 109, 137, 453,
455.
Measure, unit of, 2. 33, 34, 144, 188.
Menelaus, theorem of, 319.
Midpoint of a segment, 295, 502.
MU, 190.
Minor of a determinant, 485.
Minor axis of ellipse, 340.
Multiple roots, 460.
Multiplevalued function, 20, 265 ff.
Multiplication, 44, 50, 52, 239, 440.
Napier, J., 212.
Natural logarithms, 226.
Newton's method of approximation,
471 ff.
Nicomedes, 385.
Normal, to a curve, 359 ; — form of
the equation of a straight line, 316 ;
of a plane, 505.
Number, 2, 33 ; real, 3, 36 ; rational
and irrational, 34 ; positive and neg
ative, 3; complex (imaginary), 432 ff.
Octant, 495.
Onevalued function, 20.
Ordinate, 39.
Origin, 38, 163, 495.
Orthogonal circles, 331, 334.
Orthogonal projection in space, 494.
Parabola, 354 ff. ; definition of, 109,
268, 338; equations of, 109, 282,
354 ff.; properties of, 355, 362;
slope of, 268 ; tangents and
normals to, 359, 362 ; latus rectxim
of, 365.
Parabolic curves, 140.
Parabolic reflector, 364.
Parabolic spiral, 389.
Paraboloid, of revolution, 364 ;
elliptic, 526; hyperbolic, 527.
Parallel lines, in plane, 85 ; in space,
500.
Parameter, definition of, 90, 392;
of system of lines, 90.
Parametric equations, 392 ff.
Partial fractions, 416 ff.
Pascal, B., 383,431 ; — 's triangle, 431.
Peaucellier, inversor of, 336.
Pencil, of lines, 91; of circles, 332;
of planes, 510.
INDEX
547
Period of trigonometric functions,
157, 159, 162.
Permutations, 420 ff.
Perpendicular lines, in a plane, 86 ;
in space, 500.
Plane, 504 ff. ; vectors in a, 435.
Point of division, in a plane, 295 ;
in space, 501.
Polar and Pole, with respect to a
circle, 328 ; with respect to a
conic, 364, 373.
Polar axis, 163.
Polar coordinates, in a plane, 163 ff.,
377 ff. ; in space, 530.
Polar form of complex number,
438.
Polygon, area of, 301.
Polynomials, 402, 449 ff.
Power function, 140.
Powers, 53, 218, 444 ; table of, 534.
Prime number, 414.
Principal diagonal of a determinant,
475, 478.
Principal value of inverse trigono
metric function, 193 ff.
Probability, 426 ff.
Product formulas in trigonometry,
207.
Progression, arithmetic and geomet
ric, 216.
ProjectUe, 397 ff.
Projecting cylinder of a curve, 516.
Projection, 196 ff. ; central, 370;
orthogonal in space, 494, 497.
Pure imaginary number, 432.
Quadrant, 39 ; angles in a, 145.
Quadratic equation, 120 ff.
Quadratic function, 98 ff. ; applica
tions of, 115 ff.; slope of, 274,
287 ; implicit, 265 ff., 514 ff.
Quantity, 2, 4 ; directed, 4.
Radian, 188.
Radical axis and center of two circles,
332, 334.
Radius vector, 163, 530.
Range, of variable, 23; of a pro
jectile, 397. "■
Ratio, 33 ; simple, 294, 319.
Rational numbers, 33 ; as roots of an
equation, 467.
Reciprocal spiral, 386.
Rectangular coordinates, in a plane,
38 ff.; in space, 495.
Reference, axes of, 38.
Reflector, parabolic, 364.
Relative error, 236.
Remainder theorem, 411.
Revolution, surfaces of, 364, 519,
520.
Roots, of numbers, 444 ; table of,
534 ; of an equation, 120, 455 ff. ;
equal, of a quadratic, 124 ; rela
tion of, to coefficients, 125, 473 ;
complex, 462 ; rational, 467 ;
irrational, 470 ; multiple, 460 ;
Newton's method of approximation
to, 470 ff.
Rotation in a plane, 200.
Rounded numbers, 236 ff.
Scale, arithmetic and algebraic,
3, 5 ; rectilinear, uniform and non
uniform, 5 ; logarithmic, 252.
Secant, definition of, 168 ; graph of,
173 ; line representation of, 169.
Section of a surface, 517.
Segment, directed in a plane, 5, 48 ;
in space, 497 ; — s to represent
statistical data, 6.
Shear, 132, 288.
Significant figures, 236.
Signs, law of, 49.
Simple ratio, 294, 319.
Simultaneous equations, 94 ff., 476,
480, 490.
Sine, definition of, 147 ; variation of,
157 ; graph of, 158.
Singlevalued function, 20.
Slide rule, 252 ff.
Slope, 73, 135, 268, 273, 285, 287,
290, 449.
Solution, of quadratic equations,
120 ff. ; of algebraic equations,
468 ff. ; of trigonometric equations,
174; of exponential equations,
" 234 f of triangles, 181 ff., 244 ff.
Sphere, 514.
548
INDEX
Spherical co5rdinates, 530.
Spirals, 386 ff.
Statistical data and graphs, 6, 26.
Straight line, 64 ff., 500 ff. ; slope
of, 73; equations of, 83, 307 ff.,
511; intercept form of, 89, 315;
normal form of, 315 ; parallel and
perpendicular — s, 85, 500 ; system
of — s, 90 ; intersection of — s, 93 ;
polar equation of, 380 ; pencil of
— s, 91 ; direction cosines and angles
of, 498.
Subnormal, 364.
Successive derivatives, 458.
Sum of linear functions, 79.
Surd, 35.
Surface, 504 ff., 514 ff. ; of revolu
tion, 364, 519 ff.
Symmetric equations of straight
line, 511.
Synthetic division, 412.
Table, of squares, etc., 118, 534 ; of
logarithms, 536 ff. ; of trigono
metric functions, 538 ff.
Tabular representation of a function,
13.
Tangent (to a curve), definition of,
103 ; to a circle, 324 ff. ; to a
conic, 360 ff.; slope of, 73, 107,
135, 268, 273, 285, 287, 290, 449 ;
slope forms of equation, 359 ;
point forms of equation, 361.
Tangent (trigonometry) , definition
of, 147; variation of, 160; graph
of, 161 ; line representation of,
172 ; law of — s, 209.
Taylor's theorem, 458.
Term, 51.
Trace of a surface, 517.
Transverse axis of a hyperbola, 280,
349.
Triangle, area of, 186 ; angles of, 210 ;
solution of, 181 ff., 244 ff.
Trigonometric equations, 174.
Trigonometric functions, definitions
of, 147, 168 ; graphs of, 158, 159,
161, 167, 173, 193, 194; variation
of, 157 ff. ; computation of, 148 ff.,
152 ff. ; periods of, 157, 159, 162 ;
inverse, 192 ff. ; formulas, 179, 180,
181, 201, 202, 204, 205, 207, 446;
application of De Moivre's
theorem, to expansion of, 446 ;
logarithms of, 242, 538 ff. ; tables
of, 538 ff.
Trisection of an angle, 388.
Twovalued function, 20, 265 ff.
Variable, definition of, 12 ; independ
ent and dependent, 12 ; range of,
23.
Vector, definition of, 5, 434 ; addition
of — s, 435 ; components of,
436.
Vectorial angle, 163.
Versed sine, 168.
Vertex of a conic, 109, 343, 349.
Zero of a function. 455.
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III. The logical as well as the practical side of the subject is emphasized.
IV. The arrangement of material is pedagogical.
V. Basal theorems are printed in blackface type.
VI. The book conforms to the recommendations of the National Com
mittee on the Teaching of Geometry.
VII. Typography and binding are excellent. The latter is the reenforced
tape binding that is characteristic of Macmillan textbooks.
" Geometry is likely to remain primarily a cultural, rather than an informa
tion subject," say the authors in the preface. " But the intimate connection
of geometry with human activities is evident upon every hand, and constitutes
fully as much an integral part of the subject as does its older logical and
scholastic aspect." This connection with human activities, this application
of geometry to real human needs, is emphasized in a great variety of problems
and constructions, so that theory and application are inseparably connected
throughout the book.
These illustrations and the many others contained in the book will be seen
to cover a wider 7ange than is usual, even in books that emphasize practical
applications to a questionable extent. This results in a better appreciation
of the significance of the subject on the part of the student, in that he gains a
truer conception of the wide scope of its application.
The logical as well as the practical side of the subject is emphasized.
^ Definitions, arrangement, and method of treatment are logical. The defi
nitions are particularly simple, clear, and accurate. The traditional manner
of presentation in a logical system is preserved, with due regard for practical
applications. Proofs, both foimal and informal, are strictly logical.
THE MACMILLAN COMPANY
Publishers 64 G6 Fifth Avenue New York
Analytic Geometry and Principles of Algebra
BY
ALEXANDER ZIWET
Professor of Mathematics, the University of Michigan
AND LOUIS ALLEN HOPKINS
Instructor in Mathematics, the University of Michigan
Edited by EARLE RAYMOND HEDRICK
Cloth, via + j6g pp., appendix, answers, index, i2mo, $i.7S
This work combines with analytic geometry a number of topics traditionally
treated in college algebra that depend upon or are closely associated with
geometric sensation. Through this combination it becomes possible to show
the student more directly the meaning and the usefulness of these subjects.
The idea of coordinates is so simple that it might (and perhaps should) be
explained at the very beginning of the study of algebra and geometry. Real
analytic geometry, however, begins only when the equation in two variables
is interpreted as defining a locus. This idea must be introduced very gradu
ally, as it is difficult for the beginner to grasp. The familiar loci, straight
line and circle, are therefore treated at great length.
In the chapters on the conic sections only the most essential properties of
these curves are given in the text ; thus, poles and polars are discussed only
in connection with the circle.
The treatment of solid analytic geometry follows the more usual lines. But,
in view of the application to mechanics, the idea of the vector is given some
prominence; and the representation of a function of two variables by contour
lines as well as by a surface in space is explained and illustrated by practical
examples.
The exercises have been selected with great care in order not only to fur
nish sufficient material for practice in algebraic work but also to stimulate
independent thinking and to point out the applications of the theory to con
crete problems. The number of exercises is sufficient to allow the instructor
to make a choice.
To reduce the course presented in this book to about half its extent, the
parts of the text in small type, the chapters on soUd analytic geometry, and
the more difficult problems throughout may be omitted.
THE MACMILLAN COMPANY
Publishers 6466 Fifth Avenue New York
Elements of Analytic Geometry
BY
ALEXANDER ZIWET
Professor of Mathematics, the University of Michigan
AND LOUIS ALLEN HOPKINS
Instructor in Mathematics, the University of Michigan
Edited by EARLE RAYMOND HEDRICK
As in most colleges the course in analytic geometry is preceded by a course
in advanced algebra, it appeared desirable to publish separately those parts
of the authors' "Analytic Geometry and Principles of Algebra" which deal
with analytic geometry, omitting the sections on algebra. This is done in the
present work.
In plane analytic geometry, the idea of function is introduced as early as
possible; and curves of the form>' =/(x), where /(;c) is a simple polynomial,
are discussed even before the conic sections are treated systematically. This
makes it possible to introduce the idea of the derivative ; but the sections
dealing with the derivative may be omitted.
In the chapters on the conic sections only the most essential properties of
these curves are given in the text ; thus, poles and polars are discussed only
in connection with the circle.
The treatment of solid analytic geometry follows more the usual lines.
But, in view of the application to mechanics, the idea of the vector is given
some prominence ; and the representation of a function of two variables by
contour lines as well as by a surface in space is explained and illustrated by
practical examples.
The exercises have been selected with great care in order not only to fur
nish sufficient material for practice in algebraic work, but also to stimulate
independent thinking and to point out the applications of the theory to con
crete problems. The number of exercises is sufficient to allow the instructor
to make a choice.
THE MACMILLAN COMPANY
Publishers 6466 Fifth Avenue New York
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General Library
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